Hello, aspiring chemists! Today, we're going to unravel a fascinating connection in the world of chemical equilibrium. We've talked about equilibrium constants, right? Those magical numbers that tell us the extent to which a reaction proceeds in the forward or reverse direction. But here's the thing: sometimes we express these constants using concentrations, and sometimes using pressures. What if a reaction involves gases, and we want to switch between these two ways of expressing equilibrium? That's where our topic for today comes in: the Relationship between Kp and Kc.

Don't worry, we'll start from the very beginning, building our understanding brick by brick. By the end, you'll not only know the formula but also truly understand *why* it works!

### 1. Revisit: The Idea of Equilibrium Constants (K)

Remember, for a reversible reaction, at equilibrium, the rates of the forward and reverse reactions become equal. At this point, the concentrations of reactants and products stop changing, but the reaction is still happening in both directions! To quantify this state, we use an equilibrium constant, which is a ratio of product concentrations/pressures to reactant concentrations/pressures, each raised to the power of their stoichiometric coefficients.

Now, depending on the nature of the reactants and products, especially whether they are in solution or gas phase, we use different types of equilibrium constants.

### 2. Enter Kc: The Concentration Constant

When reactions occur in solution (like acids and bases mixing in water) or involve substances whose amounts are best described by their molar concentrations, we use Kc. The 'c' stands for concentration.

For a general reversible reaction:
aA + bB ⇌ cC + dD

The equilibrium constant Kc is expressed as:
$$ K_c = frac{[C]^c [D]^d}{[A]^a [B]^b} $$
Here, [A], [B], [C], and [D] represent the molar concentrations (moles per liter, mol/L) of the reactants and products at equilibrium. Solids and pure liquids are usually omitted from the expression because their concentrations remain essentially constant.

### 3. Meet Kp: The Pressure Constant

What about reactions involving gases? For these, it's often more convenient to measure and express their amounts in terms of their partial pressures. That's where Kp comes into play. The 'p' stands for partial pressure.

For the same general reversible reaction, but now with all components being gases:
aA(g) + bB(g) ⇌ cC(g) + dD(g)

The equilibrium constant Kp is expressed as:
$$ K_p = frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} $$
Here, P_A, P_B, P_C, and P_D represent the partial pressures (in units like atmospheres (atm) or bar, or Pascal (Pa)) of the gaseous reactants and products at equilibrium. Again, only gaseous species are included in Kp expressions.

### 4. Why Do We Need a Relationship Between Kp and Kc?

Think about it: for a reaction involving gases, we *could* describe its equilibrium using either concentrations (mol/L) or partial pressures (atm). Both are valid ways to quantify the amount of gas present. Wouldn't it be handy to have a way to convert between them? Imagine you're given Kp for a reaction, but you need Kc for a different calculation, or vice versa. This is exactly why establishing a relationship between Kp and Kc is so crucial! It allows us to seamlessly switch perspectives.

### 5. Building the Intuition: Connecting Pressure and Concentration for Gases

How are partial pressure and concentration related for gases? This is where our old friend, the Ideal Gas Law, comes to the rescue!

The Ideal Gas Law states:
PV = nRT

Where:
* P = Pressure of the gas
* V = Volume of the container
* n = Number of moles of the gas
* R = Ideal gas constant (e.g., 0.0821 L·atm/(mol·K))
* T = Absolute temperature (in Kelvin)

Let's rearrange this equation to highlight the connection:
Divide both sides by V:
$$ P = frac{n}{V} RT $$

Now, what is n/V? It's the number of moles per unit volume, which is precisely the definition of molar concentration (usually denoted by C or [ ]).

So, for any individual gas, we can write:
$$ P = CRT quad ext{or} quad P = [Gas]RT $$

This is the key! This simple equation tells us that for a gas at a constant temperature, its partial pressure is directly proportional to its molar concentration. We can use this to link Kp and Kc.

### 6. Deriving the Relationship: Step-by-Step!

Let's consider our general gas-phase equilibrium again:
aA(g) + bB(g) ⇌ cC(g) + dD(g)

We have the expressions for Kc and Kp:
$$ K_c = frac{[C]^c [D]^d}{[A]^a [B]^b} quad ext{......... (1)} $$
$$ K_p = frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} quad ext{......... (2)} $$

Now, let's use our bridge equation, P = [Gas]RT, to substitute the partial pressures in the Kp expression:

* For reactant A: $P_A = [A]RT$
* For reactant B: $P_B = [B]RT$
* For product C: $P_C = [C]RT$
* For product D: $P_D = [D]RT$

Substitute these into the Kp expression (Equation 2):
$$ K_p = frac{([C]RT)^c ([D]RT)^d}{([A]RT)^a ([B]RT)^b} $$

Let's expand the terms:
$$ K_p = frac{[C]^c (RT)^c [D]^d (RT)^d}{[A]^a (RT)^a [B]^b (RT)^b} $$

Now, gather the concentration terms and the (RT) terms separately:
$$ K_p = left( frac{[C]^c [D]^d}{[A]^a [B]^b}
ight) imes left( frac{(RT)^c (RT)^d}{(RT)^a (RT)^b}
ight) $$

Look closely at the first bracketed term! Does it look familiar? Yes, it's exactly our expression for Kc (Equation 1)!

So, we can replace that part with Kc:
$$ K_p = K_c imes left( frac{(RT)^{c+d}}{(RT)^{a+b}}
ight) $$

Using the rules of exponents (x^m / x^n = x^(m-n)):
$$ K_p = K_c imes (RT)^{(c+d) - (a+b)} $$

Let's define a new term, Δn (pronounced "delta n"):
Δn = (c + d) - (a + b)

What does Δn represent? It's the sum of the stoichiometric coefficients of the gaseous products minus the sum of the stoichiometric coefficients of the gaseous reactants. In simpler words, it's the change in the number of moles of gas during the reaction.

Finally, we arrive at the beautiful relationship:
$$ mathbf{K_p = K_c (RT)^{Delta n}} $$

This is the central formula we were aiming for!

### 7. Understanding Δn (Delta n) in Detail

Crucial Point for JEE/NEET & CBSE: When calculating Δn, you must only consider the moles of *gaseous* reactants and products. Solids, liquids, and aqueous solutions are *not* included in the calculation of Δn because their concentrations (and thus partial pressures for liquids/solids, which are negligible) don't change in the same way as gases.

Let's break down Δn further:
* Δn = (moles of gaseous products) - (moles of gaseous reactants)

Example 1: Δn = 0
Consider the reaction:
H₂(g) + I₂(g) ⇌ 2HI(g)

* Moles of gaseous products = 2 (from 2HI)
* Moles of gaseous reactants = 1 (from H₂) + 1 (from I₂) = 2
* Δn = 2 - 2 = 0

In this case, Kp = Kc (RT)⁰ = Kc * 1 = Kp = Kc.
This means if there's no change in the total number of gas moles during the reaction, Kp and Kc are numerically equal.

Example 2: Δn > 0
Consider the dissociation of PCl₅:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

* Moles of gaseous products = 1 (from PCl₃) + 1 (from Cl₂) = 2
* Moles of gaseous reactants = 1 (from PCl₅)
* Δn = 2 - 1 = 1

Here, Kp = Kc (RT)¹ = Kp = Kc RT. Since RT is usually a positive value (R is positive, T in Kelvin is positive), if Δn > 0, then Kp will generally be greater than Kc (assuming RT > 1, which is often the case at typical temperatures).

Example 3: Δn < 0
Consider the synthesis of ammonia (Haber process):
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

* Moles of gaseous products = 2 (from 2NH₃)
* Moles of gaseous reactants = 1 (from N₂) + 3 (from H₂) = 4
* Δn = 2 - 4 = -2

Here, Kp = Kc (RT)⁻² = Kp = Kc / (RT)². In this scenario, Kp will be smaller than Kc.

Example 4: Including non-gaseous species
Consider the decomposition of calcium carbonate:
CaCO₃(s) ⇌ CaO(s) + CO₂(g)

* Moles of gaseous products = 1 (from CO₂)
* Moles of gaseous reactants = 0 (CaCO₃ is solid, so we don't count it)
* Δn = 1 - 0 = 1

Here, Kp = Kc (RT)¹. Remember, only gaseous moles contribute to Δn! This is a common pitfall.

### 8. Practical Considerations and JEE Focus

* Units of R: The value of R you use must match the units of pressure and volume.
* If pressure is in atm and volume in L, use R = 0.0821 L·atm/(mol·K).
* If pressure is in Pa and volume in m³, use R = 8.314 J/(mol·K) (which is also L·kPa/(mol·K)).
* Temperature (T): Always use temperature in Kelvin (K). Convert Celsius to Kelvin by adding 273.15 (or 273 for most calculations).
* Significance: This relationship is incredibly useful for interconverting Kp and Kc, especially in JEE problems where one value might be given, and the other needs to be calculated.
* JEE Advanced Tip: Sometimes, problems might involve situations where the ideal gas law doesn't perfectly apply, or they might ask you to justify the approximation. However, for most standard problems, the ideal gas law is assumed.

So there you have it! From understanding why we have different equilibrium constants to deriving the exact relationship between them using the Ideal Gas Law, we've covered the fundamentals of Kp and Kc. This formula is a powerful tool in your equilibrium arsenal, allowing you to navigate various problems with confidence. Keep practicing with different reaction examples, and you'll master it in no time!

Welcome, aspiring chemists! Today, we're going to dive deep into one of the most fundamental and frequently tested concepts in chemical equilibrium: the Relationship between Kp and Kc. You might have heard these terms before – Kp and Kc – as equilibrium constants. But why do we have two, and how are they connected? Let's unravel this mystery step-by-step, building a strong conceptual foundation that will serve you well in both board exams and competitive tests like JEE.

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### Understanding the Equilibrium Constants: Kp and Kc

Before we connect them, let's briefly recap what Kp and Kc represent individually.

#### 1. Kc: The Equilibrium Constant in terms of Molar Concentrations

The equilibrium constant, Kc, expresses the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to the power of its stoichiometric coefficient. It is primarily used for reactions occurring in aqueous solutions or when the concentrations of gaseous reactants/products are directly given.

For a general reversible reaction:

Where a, b, c, d are the stoichiometric coefficients, and A, B, C, D are the chemical species.

The expression for Kc is:

Here, [ ] denotes the molar concentration (moles/liter or Molarity) of the respective species at equilibrium. Kc is temperature-dependent and has units that vary depending on the stoichiometry of the reaction, though it's often treated as dimensionless in advanced discussions.

#### 2. Kp: The Equilibrium Constant in terms of Partial Pressures

When we deal with gaseous reactions, it's often more convenient or direct to measure the partial pressures of the reactants and products rather than their molar concentrations. This is where Kp comes into play. Kp is the equilibrium constant expressed in terms of the partial pressures of the gaseous reactants and products.

For the same general gaseous reaction:

The expression for Kp is:

Here, P denotes the partial pressure of the respective gaseous species at equilibrium, usually in atmospheres (atm) or bar. Like Kc, Kp is also temperature-dependent, and its units depend on the reaction stoichiometry.

### The Need for a Relationship: Bridging the Gap

Imagine you have experimental data from a gaseous reaction. Sometimes you measure concentrations, other times you measure partial pressures. Or perhaps you have a value for Kc and need to calculate Kp for a specific application. This is why a relationship between these two constants is absolutely essential. It allows us to convert between concentration and pressure terms for gaseous species, ensuring consistency and flexibility in our calculations.

### Derivation of the Relationship between Kp and Kc

Let's derive the connection using the Ideal Gas Law. This is a crucial derivation for your understanding and will solidify the concept.

Consider a general reversible gaseous reaction:

We know the expression for Kp is:

And the expression for Kc is:


Now, let's bring in the Ideal Gas Law:

Where:
* P = Pressure
* V = Volume
* n = Number of moles
* R = Ideal gas constant
* T = Absolute temperature (in Kelvin)

From the Ideal Gas Law, we can express pressure in terms of concentration:

We know that molar concentration (C or [ ]) = n/V.
So, for any gaseous species 'X', its partial pressure P_X can be written as:

Now, let's substitute this relationship for each partial pressure in the Kp expression (Equation 1):

* P_A = [A]RT
* P_B = [B]RT
* P_C = [C]RT
* P_D = [D]RT

Substituting these into the Kp expression:

Let's rearrange the terms:

Group the concentration terms and the (RT) terms:

Notice that the first part of this expression is simply Kc (from Equation 2).
So, we can write:

Using the rule of exponents (x^m / xⁿ = x^m-n), we get:

Let's define Δn_g (delta n gas) as the change in the number of moles of gaseous species during the reaction.

In our general reaction:

Substituting Δn_g into our derived equation, we finally get the highly important relationship:

This is the fundamental equation you need to remember and understand!

### Decoding Δn_g: The Change in Moles of Gaseous Species

The term Δn_g is absolutely critical. Remember these key points:
1. Only Gaseous Species: When calculating Δn_g, you must only consider the moles of gaseous reactants and products. Solids and liquids are excluded because their concentrations (and hence partial pressures) do not change significantly and are typically incorporated into the value of K.
2. Product Moles - Reactant Moles: It's always (moles of gaseous products) minus (moles of gaseous reactants).

Let's look at how Δn_g affects the relationship:

#### Case 1: Δn_g = 0
If the sum of the stoichiometric coefficients of gaseous products equals the sum of the stoichiometric coefficients of gaseous reactants, then Δn_g = 0.
In this case, (RT)⁰ = 1.
So, the relationship simplifies to:

Example:

* Moles of gaseous products = 2 (for HI)
* Moles of gaseous reactants = 1 (for H₂) + 1 (for I₂) = 2
* Δn_g = 2 - 2 = 0
Therefore, for this reaction, Kp = Kc.

#### Case 2: Δn_g > 0
If the sum of the stoichiometric coefficients of gaseous products is greater than that of gaseous reactants, then Δn_g is positive.
In this case, Kp will be greater than Kc (assuming RT > 1, which is typically true at normal temperatures).

Example:

* Moles of gaseous products = 1 (for PCl₃) + 1 (for Cl₂) = 2
* Moles of gaseous reactants = 1 (for PCl₅)
* Δn_g = 2 - 1 = +1
Therefore, for this reaction, Kp = Kc(RT)¹ or Kp = KcRT.

#### Case 3: Δn_g < 0
If the sum of the stoichiometric coefficients of gaseous products is less than that of gaseous reactants, then Δn_g is negative.
In this case, Kp will be less than Kc (assuming RT > 1).

Example:

* Moles of gaseous products = 2 (for NH₃)
* Moles of gaseous reactants = 1 (for N₂) + 3 (for H₂) = 4
* Δn_g = 2 - 4 = -2
Therefore, for this reaction, Kp = Kc(RT)^-2 or Kp = Kc / (RT)².

### The Gas Constant (R) and Temperature (T)

The values of R and T are crucial in using the Kp-Kc relationship:
* T: Must always be in Kelvin (K). If given in Celsius, convert it: T(K) = T(°C) + 273.15.
* R: The choice of R depends on the units of pressure used in Kp.
* If partial pressures are in atmospheres (atm), use R = 0.0821 L atm mol⁻¹ K⁻¹. This is the most common R value in Kp-Kc calculations for JEE.
* If partial pressures are in Pascals (Pa), use R = 8.314 J mol⁻¹ K⁻¹ (which is also 8.314 Pa m³ mol⁻¹ K⁻¹).
* If partial pressures are in bar, use R = 0.08314 L bar mol⁻¹ K⁻¹.

JEE FOCUS: Always double-check the units of pressure given in the problem and choose the appropriate value of R. Mismatched units are a common source of errors!

### Advanced Considerations and JEE Insights

1. Heterogeneous Equilibria: When solids or liquids are involved in the equilibrium, they do not appear in the Kp or Kc expressions, and therefore, they do not contribute to Δn_g.
Example:

Here, Kp = P_CO2 and Kc = [CO₂].
* Moles of gaseous products = 1 (for CO₂)
* Moles of gaseous reactants = 0 (CaCO₃ is solid)
* Δn_g = 1 - 0 = +1
So, Kp = Kc(RT)¹ or Kp = KcRT. This holds true as P_CO2 = [CO₂]RT.

2. Units of Kp and Kc (Revisited): While Kp and Kc are often quoted as dimensionless, their true units depend on Δn_g.
* If Δn_g = 0, Kp and Kc are dimensionless.
* If Δn_g = +1, Kp has units of (atm) and Kc has units of (mol/L). The (RT) term (L atm mol⁻¹ K⁻¹ * K = L atm mol⁻¹) makes them consistent.
* The relationship Kp = Kc(RT)^Δn_g inherently balances the units.

3. Temperature Dependence: Both Kp and Kc are constants for a given reaction at a specific temperature. The (RT)^Δn_g term in the relationship explicitly shows that the ratio Kp/Kc is also temperature-dependent.

### Worked Examples

Let's solidify our understanding with some practical applications.

#### Example 1: Calculate Kp from Kc
For the reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
Given Kc = 2.8 x 10² at 1000 K. Calculate Kp. (R = 0.0821 L atm mol⁻¹ K⁻¹)

Step-by-step Solution:
1. Identify gaseous species and their stoichiometric coefficients:
* Products: SO₃ (2 moles, gas)
* Reactants: SO₂ (2 moles, gas), O₂ (1 mole, gas)
2. Calculate Δn_g:
Δn_g = (Moles of gaseous products) - (Moles of gaseous reactants)
Δn_g = (2) - (2 + 1) = 2 - 3 = -1
3. Apply the Kp-Kc relationship:
Kp = Kc (RT)^Δn_g
Kp = (2.8 x 10²) * (0.0821 * 1000)^-1
4. Calculate the (RT) term:
RT = 0.0821 L atm mol⁻¹ K⁻¹ * 1000 K = 82.1 L atm mol⁻¹
5. Substitute and solve for Kp:
Kp = (2.8 x 10²) * (82.1)^-1
Kp = 280 / 82.1
Kp ≈ 3.41

#### Example 2: Calculate Kc from Kp (Heterogeneous Equilibrium)
For the reaction: C(s) + H₂O(g) ⇌ CO(g) + H₂(g)
Given Kp = 3.25 atm at 1200 K. Calculate Kc. (R = 0.0821 L atm mol⁻¹ K⁻¹)

Step-by-step Solution:
1. Identify gaseous species and their stoichiometric coefficients:
* Products: CO (1 mole, gas), H₂ (1 mole, gas)
* Reactants: H₂O (1 mole, gas), C (solid - DO NOT INCLUDE in Δn_g)
2. Calculate Δn_g:
Δn_g = (Moles of gaseous products) - (Moles of gaseous reactants)
Δn_g = (1 + 1) - (1) = 2 - 1 = +1
3. Apply the Kp-Kc relationship:
Kp = Kc (RT)^Δn_g
3.25 = Kc * (0.0821 * 1200)⁺¹
4. Calculate the (RT) term:
RT = 0.0821 L atm mol⁻¹ K⁻¹ * 1200 K = 98.52 L atm mol⁻¹
5. Substitute and solve for Kc:
3.25 = Kc * 98.52
Kc = 3.25 / 98.52
Kc ≈ 0.0330

#### Example 3: When Δn_g = 0
For the reaction: H₂(g) + I₂(g) ⇌ 2HI(g)
If Kc = 54.3 at 700 K, what is Kp?

Step-by-step Solution:
1. Identify gaseous species and their stoichiometric coefficients:
* Products: HI (2 moles, gas)
* Reactants: H₂ (1 mole, gas), I₂ (1 mole, gas)
2. Calculate Δn_g:
Δn_g = (2) - (1 + 1) = 2 - 2 = 0
3. Apply the Kp-Kc relationship:
Kp = Kc (RT)^Δn_g
Kp = Kc (RT)⁰
Kp = Kc * 1
Kp = Kc = 54.3

This shows that when Δn_g is zero, the values of Kp and Kc are identical, regardless of temperature or the value of R.

### Conclusion

The relationship Kp = Kc (RT)^Δn_g is a cornerstone of chemical equilibrium calculations for gaseous systems. It's not just a formula to memorize, but a derived consequence of the Ideal Gas Law connecting concentration and pressure. Mastering its application, especially the correct calculation of Δn_g and the selection of R and T units, is vital for success in your chemistry studies and competitive exams. Remember, careful unit handling and understanding the underlying principles will always lead you to the correct answer!

The relationship between the equilibrium constants Kp (in terms of partial pressures) and Kc (in terms of molar concentrations) is a fundamental concept in chemical equilibrium. Remembering the formula and its components accurately is crucial for JEE and board exams.

Here are some effective mnemonics and shortcuts to help you recall this relationship easily:

### 1. Main Formula Mnemonic: Kp = Kc(RT)^Δn

This is the core relationship you need to remember.

* Formula: Kp = Kc (RT)^Δn
* Kp: Equilibrium constant in terms of partial pressures.
* Kc: Equilibrium constant in terms of molar concentrations.
* R: Ideal gas constant (usually 0.0821 L atm mol⁻¹ K⁻¹ for Kp in atm, or 8.314 J mol⁻¹ K⁻¹ if using SI units).
* T: Absolute temperature in Kelvin.
* Δn: Change in the number of moles of gaseous products and reactants.

* Mnemonic: "King Penguins Keep Cool (by) Running To Den."
* K P maps to K P
* K C maps to K C
* R T maps to R T
* Den maps to Δn (Delta N)
* This phrase directly helps you reconstruct the formula: K P = K C (R T)^Δn.

### 2. Calculating Δn Mnemonic: (Moles of Gaseous Products) - (Moles of Gaseous Reactants)

The calculation of Δn is a common point of error. Remember these two key rules:

* Rule 1: Only Gaseous Species: Δn considers only the stoichiometric coefficients of *gaseous* products and *gaseous* reactants. Solids and liquids are explicitly excluded.
* Rule 2: Products Minus Reactants: It's always the moles of products minus the moles of reactants.

* Mnemonic: "Δn is for Gases Only: Products Minus Reactants."
* Gases Only: Reminds you to ignore solids and liquids.
* Products Minus Reactants: Gives the correct order for subtraction.

### 3. Special Case Mnemonic: When Δn = 0

This is a frequently tested scenario where the relationship simplifies.

* Condition: If Δn = 0, then (RT)⁰ = 1.
* Result: Kp = Kc (since Kp = Kc * 1)

* Mnemonic: "Delta N Zero means K Pays K Cash!"
* D N Zero: When Delta N is zero.
* K Pays K Cash: Kp equals Kc.
* This simple phrase helps you recall that when Δn is zero, Kp and Kc become equal.

### Quick Tips for Exam Success:

* Units of R: Always ensure you use the correct value of R. If partial pressures are in atmospheres, use R = 0.0821 L atm mol⁻¹ K⁻¹.
* Temperature in Kelvin: Temperature (T) must always be in Kelvin. Convert Celsius to Kelvin by adding 273.15 (or 273 for quick calculations).
* JEE vs. CBSE: This relationship is critical for both. JEE might involve more complex Δn calculations or scenarios, while CBSE will test direct application of the formula.

Mastering these mnemonics will ensure you accurately recall and apply the Kp and Kc relationship, saving valuable time and preventing common errors in your exams!

Quick Tips: Relationship between Kp and Kc

Understanding the relationship between Kp and Kc is crucial for solving equilibrium problems, especially in JEE Main. Here are some quick tips to master this concept:

1. The Core Formula:

The fundamental relationship is:

Kp = Kc (RT)Δng

Where:

• Kp: Equilibrium constant in terms of partial pressures.


• Kc: Equilibrium constant in terms of molar concentrations.


• R: Universal gas constant.


• T: Absolute temperature in Kelvin.


• Δng: Change in the number of moles of gaseous products and reactants.

2. Decoding Δng (The Most Common Error Point):

This is where most students make mistakes.

• Definition: Δng = (Sum of stoichiometric coefficients of gaseous products) - (Sum of stoichiometric coefficients of gaseous reactants)


• Crucial Point: Only consider gaseous species! Solids and liquids do not contribute to Δng. Their concentrations (or partial pressures) are considered constant and are absorbed into the equilibrium constant.


• Example: For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
  
  
  
  • Moles of gaseous products = 2 (for NH₃)
  
  
  • Moles of gaseous reactants = 1 (for N₂) + 3 (for H₂) = 4
  
  
  • Δng = 2 - 4 = -2
  
  
  
  


• Example with heterogeneous equilibrium: For CaCO₃(s) ⇌ CaO(s) + CO₂(g)
  
  
  
  • Moles of gaseous products = 1 (for CO₂)
  
  
  • Moles of gaseous reactants = 0 (CaCO₃ is solid)
  
  
  • Δng = 1 - 0 = 1
  
  
  
  

3. Value of R and Units:

The value of R depends on the units of pressure (for Kp) and volume (for concentration terms in Kc).

• For calculations involving Kp and Kc, the most common value to use is R = 0.0821 L atm mol⁻¹ K⁻¹.
  
  This implies pressure is in atmospheres and volume for concentration is in Liters.


• Ensure the temperature T is always in Kelvin. (T in °C + 273.15).

4. Special Cases of Δng:

• If Δng = 0: Then (RT)Δng = (RT)⁰ = 1.
  
  In this case, Kp = Kc. This is a common scenario in many reactions (e.g., H₂(g) + I₂(g) ⇌ 2HI(g)).


• If Δng > 0 (positive): Then Kp > Kc. (Assuming RT > 1, which is always true for T > 0 K)


• If Δng < 0 (negative): Then Kp < Kc. (Assuming RT > 1)

5. JEE Specific Tip:

For JEE Main, meticulous calculation of Δng is key. Always write down the balanced chemical equation and clearly identify the gaseous species before calculating Δng. A sign error or inclusion of solids/liquids will lead to a completely wrong answer.

Stay focused and practice calculating Δng for various types of reactions!

Intuitive Understanding: Relationship between Kp and Kc

Understanding the relationship between the equilibrium constants Kp and Kc is crucial for both conceptual clarity and problem-solving in chemical equilibrium. While both constants describe the same state of equilibrium for a reversible reaction, they do so using different measures of reactant and product amounts.

What are Kp and Kc?

• 
  Kc (Equilibrium Constant in terms of Concentrations): This constant uses the molar concentrations (moles/liter) of reactants and products at equilibrium. It is primarily used when dealing with reactions involving species in aqueous solutions or when concentrations are the most convenient measure.
  


• 
  Kp (Equilibrium Constant in terms of Partial Pressures): This constant uses the partial pressures of gaseous reactants and products at equilibrium. It is exclusively used for reactions involving gaseous species, as pressure is a direct and convenient measure for gases.
  

The Core Intuitive Link: Ideal Gas Law

The fundamental reason Kp and Kc are related stems directly from the Ideal Gas Law. For an ideal gas:

PV = nRT

Where:

• P = Partial pressure of the gas


• V = Volume of the container


• n = Moles of the gas


• R = Ideal gas constant


• T = Absolute temperature (in Kelvin)

Rearranging this equation, we get:

P = (n/V)RT

Since (n/V) represents the molar concentration (C) of the gas, we can write:

P = CRT

This equation is the key: it shows that for any gaseous species at a given temperature, its partial pressure (P) is directly proportional to its molar concentration (C). The 'RT' factor is the proportionality constant.

How Δn_g Comes into Play

Consider a general reversible gaseous reaction:

aA(g) + bB(g) ↔ cC(g) + dD(g)

Kc = [C]^c[D]^d / [A]^a[B]^b

Kp = (P_C)^c(P_D)^d / (P_A)^a(P_B)^b

Now, substitute P = CRT for each gaseous species into the Kp expression:

Kp = ([C]RT)^c([D]RT)^d / ([A]RT)^a([B]RT)^b

Kp = ([C]^c[D]^d / [A]^a[B]^b) * (RT)^c+d / (RT)^a+b

We recognize the first part as Kc. The exponent (c+d) is the sum of moles of gaseous products, and (a+b) is the sum of moles of gaseous reactants.
Let Δn_g = (moles of gaseous products) - (moles of gaseous reactants) = (c+d) - (a+b).
So, the relationship simplifies to:

Kp = Kc(RT)^Δn_g

Intuitive Meaning of Δn_g

• 
  If Δn_g = 0: This means the total number of moles of gaseous products equals the total number of moles of gaseous reactants. In this scenario, the (RT)^Δn_g term becomes (RT)⁰ = 1. Therefore, Kp = Kc. Intuitively, if there's no net change in the number of gas particles, the equilibrium constant measured by pressure (Kp) will be numerically identical to the one measured by concentration (Kc), as the pressure-concentration proportionality factor (RT) cancels out perfectly.
  


• 
  If Δn_g ≠ 0: There is a net change in the number of gas moles. The (RT)^Δn_g factor accounts for this difference.
  
  
  
  • If Δn_g > 0 (more gaseous products): Kp will be larger than Kc (since RT is usually > 1). More gas moles mean a greater pressure contribution relative to concentration.
  
  
  • If Δn_g < 0 (fewer gaseous products): Kp will be smaller than Kc. Fewer gas moles mean a smaller pressure contribution relative to concentration.
  
  
  
  

JEE & CBSE Tip: Remember that Δn_g considers only gaseous species. Solids and liquids are excluded from the calculation of Δn_g as their concentrations/partial pressures do not change significantly and are incorporated into the constant itself or are not relevant for Kp calculations. Always use R = 0.0821 L atm mol^-1 K^-1 when pressures are in atmospheres, and T in Kelvin.

Understanding the relationship between Kp and Kc, represented by the equation Kp = Kc (RT)^Δn_g, can be simplified by drawing analogies to common concepts. Both Kp and Kc quantify the extent of a reaction at equilibrium, but they do so using different units or 'perspectives'.

Analogy 1: Currency Exchange Rates

• Imagine a product (your equilibrium state) whose value can be expressed in two different currencies: "Concentration Currency" (Kc) and "Pressure Currency" (Kp).


• Both currencies represent the same fundamental value, but their numerical figures might differ based on the exchange rate.


• The term (RT)^Δn_g acts as the dynamic exchange rate between these two currencies.


• R and T are like universal factors influencing the exchange (e.g., global economic conditions, fixed bank rates).


• Δn_g is the crucial component here:
  
  
  
  • If Δn_g = 0 (no net change in moles of gaseous substances), it's like the exchange rate is 1:1. One unit of "Concentration Currency" is exactly equal to one unit of "Pressure Currency" (Kp = Kc).
  
  
  • If Δn_g ≠ 0, then there's a specific conversion factor needed, just like converting USD to EUR – the numerical value changes based on the rate.
  
  
  
  


• This analogy helps to visualize that Kp and Kc describe the same equilibrium point but in different 'monetary terms', requiring a specific conversion factor when the 'basis' (moles of gas) changes.

Analogy 2: Measuring "Impact" vs. "Quantity"

• Consider measuring the 'strength' or 'extent' of a chemical reaction at equilibrium.


• Kc is like measuring the 'strength' based on the actual quantities or molar amounts (concentrations) of reactants and products present. It's a direct count or density measurement.


• Kp is like measuring the 'strength' based on the force or impact these gaseous substances exert (partial pressures). It's about the 'push' they create.


• While both measure the same underlying 'strength' of the equilibrium, the medium (gaseous phase) means that the 'impact' (pressure) can change disproportionately to the 'quantity' (concentration) if the total number of gas molecules changes.


• The term (RT)^Δn_g serves as the conversion bridge that accounts for this difference:
  
  
  
  • If Δn_g = 0, the number of gas particles remains constant. Thus, the 'quantity' and 'impact' are directly proportional, leading to Kp = Kc.
  
  
  • If Δn_g ≠ 0, a change in the number of gaseous moles will directly affect the total pressure exerted for a given concentration, hence requiring the (RT)^Δn_g factor to accurately convert between 'quantity-based' (Kc) and 'impact-based' (Kp) equilibrium constants.
  
  
  
  

Key Takeaway for JEE/CBSE: These analogies reinforce that Kp and Kc are fundamentally describing the same chemical equilibrium. The equation Kp = Kc (RT)^Δn_g is merely a mathematical tool to convert between two different ways of expressing this equilibrium constant, particularly essential when gaseous components are involved and their total moles change during the reaction (Δn_g ≠ 0).

To effectively understand the relationship between Kp and Kc, it is crucial to have a strong grasp of several foundational concepts from Chemical Equilibrium and States of Matter. Mastering these prerequisites will ensure a smooth learning curve for this advanced topic.

Here are the essential prerequisites:

• 1. Chemical Equilibrium Basics:
  
  
  
  • Reversible Reactions: Understanding that reactions can proceed in both forward and reverse directions.
  
  
  • Dynamic Equilibrium: Knowing that at equilibrium, the rates of forward and reverse reactions are equal, leading to constant macroscopic properties.
  
  
  • Law of Mass Action: The ability to write the general expression for the equilibrium constant based on the stoichiometry of a balanced chemical equation.
  
  
  
  

  Why it's important: Kp and Kc are specific types of equilibrium constants. Without understanding what equilibrium is and how constants are derived, the relationship will lack context. This is fundamental for both CBSE and JEE.

  
  



• 2. Equilibrium Constant in Terms of Concentration (Kc):
  
  
  
  • Definition: Expressing the equilibrium constant for a reaction involving species in solution or gas phase, using their molar concentrations (moles/liter).
  
  
  • Calculation: Ability to calculate Kc from given equilibrium concentrations.
  
  
  
  

  Why it's important: Kc is one of the two variables whose relationship is being established. A clear understanding of its formulation is non-negotiable for both CBSE and JEE.

  
  



• 3. Partial Pressures of Gases:
  
  
  
  • Dalton's Law of Partial Pressures: Understanding that the total pressure exerted by a mixture of non-reacting gases is the sum of the partial pressures of individual gases.
  
  
  • Partial Pressure Definition: Knowing that the partial pressure of a gas in a mixture is the pressure it would exert if it alone occupied the same volume at the same temperature.
  
  
  
  

  Why it's important: Kp is defined using partial pressures. A solid understanding of how partial pressures are calculated and interpreted is crucial, especially for JEE where more complex gas mixtures might be involved.

  
  



• 4. Ideal Gas Equation (PV=nRT):
  
  
  
  • Fundamentals: Basic understanding of the Ideal Gas Law and the meaning of each term (P, V, n, R, T).
  
  
  • Rearrangement: Ability to rearrange the equation to relate pressure (P) to molar concentration (n/V). Specifically, P = (n/V)RT = CRT, where C is molar concentration.
  
  
  
  

  Why it's important: This equation is the lynchpin connecting pressure and concentration, forming the mathematical bridge between Kp and Kc. This is a critical concept for both CBSE and JEE.

  
  



• 5. Stoichiometry and Mole Concept:
  
  
  
  • Balanced Chemical Equations: The ability to balance chemical equations correctly.
  
  
  • Mole Ratios: Understanding how mole ratios from balanced equations relate to changes in moles of reactants and products.
  
  
  
  

  Why it's important: The term Δn_g in the Kp-Kc relationship directly arises from the stoichiometry of gaseous reactants and products. Errors in balancing equations will lead to incorrect Δn_g values, impacting the final result. Essential for both CBSE and JEE.

  
  

Ensure you are comfortable with these topics before delving into the derivation and application of the Kp-Kc relationship. A little time spent on these basics will save a lot of confusion later!

Understanding the relationship between K_p and K_c is a cornerstone of chemical equilibrium. To truly master this concept and apply it effectively in exams, a firm grasp of several prerequisite and related topics is essential. These topics provide the foundational knowledge and context for deriving and utilizing the K_p-K_c equation.

Key Related Topics

• 
  The Ideal Gas Equation (PV = nRT):
  
  
  
  • This is arguably the most crucial related topic, as it forms the mathematical bridge between partial pressures (P) and molar concentrations (n/V).
  
  
  • Recall that concentration, C = n/V. Substituting n/V from PV=nRT gives P = (n/V)RT = CRT.
  
  
  • This direct relationship allows for the conversion between terms involving pressure and those involving concentration, which is fundamental to deriving the K_p-K_c expression.
  
  
  • JEE Pointer: Questions often test your ability to correctly use R (gas constant) and T (temperature in Kelvin) in appropriate units.
  
  
  
  


• 
  Definition and Expression of K_c (Equilibrium Constant in terms of Molar Concentrations):
  
  
  
  • K_c expresses the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of its stoichiometric coefficient.
  
  
  • Example: For aA + bB ⇌ cC + dD, K_c = [C]^c[D]^d / [A]^a[B]^b.
  
  
  • Understanding how to correctly write K_c expressions, especially for heterogeneous equilibria where solids and pure liquids are excluded, is a prerequisite.
  
  
  
  


• 
  Definition and Expression of K_p (Equilibrium Constant in terms of Partial Pressures):
  
  
  
  • K_p expresses the ratio of the partial pressures of products to reactants at equilibrium, each raised to the power of its stoichiometric coefficient.
  
  
  • Example: For aA(g) + bB(g) ⇌ cC(g) + dD(g), K_p = (P_C)^c(P_D)^d / (P_A)^a(P_B)^b.
  
  
  • Only gaseous species are included in K_p expressions.
  
  
  
  


• 
  Stoichiometry and Balanced Chemical Equations:
  
  
  
  • The derivation of K_p = K_c(RT)^Δn_g directly depends on the change in the number of moles of gaseous species (Δn_g).
  
  
  • Δn_g = (Sum of stoichiometric coefficients of gaseous products) - (Sum of stoichiometric coefficients of gaseous reactants).
  
  
  • Correctly calculating Δn_g requires a perfectly balanced chemical equation and careful identification of gaseous components. An error in balancing or identifying phases will lead to an incorrect Δn_g and thus an incorrect relationship.
  
  
  
  


• 
  Units of K_p and K_c:
  
  
  
  • While equilibrium constants are often treated as dimensionless in advanced contexts, for problem-solving in JEE and CBSE, knowing how their units arise from the expressions is important, especially when dealing with Δn_g ≠ 0.
  
  
  • Units of K_c: (mol/L)^Δn_g
  
  
  • Units of K_p: (atm or bar)^Δn_g
  
  
  
  

Mastering these related topics will not only solidify your understanding of the K_p-K_c relationship but also enable you to tackle complex equilibrium problems with confidence. Keep practicing the identification of gaseous species and the calculation of Δn_g!

The relationship between the pressure equilibrium constant (K_p) and the concentration equilibrium constant (K_c) is fundamental in Chemical Equilibrium and a frequent source of tricky questions in both CBSE and JEE exams. The core equation is K_p = K_c(RT)^Δn. While seemingly straightforward, several common pitfalls can lead to incorrect answers.

Here are the common exam traps students fall into when dealing with K_p and K_c:

• 
  Incorrect Value or Units of R:
  
  
  
  • This is a very frequent mistake. The Gas Constant 'R' must be chosen carefully based on the units of pressure used.
  
  
  • If pressure is in atmospheres (atm), use R = 0.0821 L atm mol^-1 K^-1.
  
  
  • If pressure is in Pascal (Pa) and volume in m³, or if other SI units are used, then use R = 8.314 J mol^-1 K^-1.
  
  
  • JEE Tip: Questions often provide pressure in bar or kPa. Remember to convert them to atmospheres if using R = 0.0821, or be consistent with SI units and use R = 8.314.
  
  
  
  


• 
  Temperature in Celsius Instead of Kelvin:
  
  
  
  • The temperature 'T' in the K_p-K_c relationship (and all gas law calculations) must always be in Kelvin (K).
  
  
  • Students often forget to convert Celsius (°C) to Kelvin (K), leading to incorrect (often negative or wildly off) values. Remember: T(K) = T(°C) + 273.15 (or 273 for simpler calculations).
  
  
  
  


• 
  Errors in Calculating Δn (Delta n):
  
  
  
  • Δn = (sum of stoichiometric coefficients of gaseous products) - (sum of stoichiometric coefficients of gaseous reactants).
  
  
  • Including Non-Gaseous Species: This is arguably the most critical and common error. Δn only accounts for the moles of *gaseous* species. Solids, liquids, and aqueous solutions are NOT included in the calculation of Δn. Forgetting this will invariably lead to a wrong answer.
  
  
  • Sign Error: Ensure the order of subtraction is correct (products minus reactants). A positive Δn means K_p > K_c, while a negative Δn means K_p < K_c.
  
  
  • Ignoring Stoichiometric Coefficients: Forgetting to multiply the number of moles by their respective stoichiometric coefficients in the balanced chemical equation.
  
  
  • Example of Δn calculation: For the reaction 2SO₂(g) + O₂(g) ↔ 2SO₃(g), Δn = (moles of gaseous products) - (moles of gaseous reactants) = 2 - (2 + 1) = 2 - 3 = -1.
  
  
  • Example with heterogeneous equilibrium: For CaCO₃(s) ↔ CaO(s) + CO₂(g), Δn = (moles of gaseous products) - (moles of gaseous reactants) = 1 - 0 = 1. (Solids are excluded).
  
  
  
  


• 
  Misconception about K_p = K_c:
  
  
  
  • K_p is equal to K_c only if Δn = 0.
  
  
  • Students sometimes assume they are always equal, which is incorrect. Only when the total number of moles of gaseous products equals the total number of moles of gaseous reactants does Δn become zero, simplifying the relationship to K_p = K_c.
  
  
  
  


• 
  Unit Inconsistency:
  
  
  
  • While K_p and K_c are often treated as dimensionless quantities in calculations, their underlying units depend on Δn. Pay attention to the units of R, T, and pressure to ensure consistency.
  
  
  
  

Key takeaway: Always double-check your value for R, ensure temperature is in Kelvin, and meticulously calculate Δn, paying special attention to only include gaseous species and their correct stoichiometric coefficients. These checks can prevent most errors.

Understanding the relationship between Kp and Kc is fundamental for solving equilibrium problems involving gaseous reactions in both CBSE and JEE Main examinations. These key takeaways will summarize the essential points you need to master.

Key Relationship Formula

The equilibrium constant Kp (expressed in terms of partial pressures) and Kc (expressed in terms of molar concentrations) for a general reversible gaseous reaction are related by the following equation:

• 
  Kp = Kc(RT)^Δn_g
  

Understanding Each Term

It is crucial to know what each term in the equation represents and its correct application:

• Kp: Equilibrium constant expressed in terms of partial pressures of gaseous reactants and products. Applicable only for reactions involving gases.


• Kc: Equilibrium constant expressed in terms of molar concentrations (mol/L) of reactants and products. Applicable to all types of reactions (gaseous, aqueous, etc.).


• R: The Universal Gas Constant. Its value depends on the units of pressure and volume used.
  
  
  
  • For Kp-Kc relationship calculations, the most commonly used value is R = 0.0821 L atm mol^-1 K^-1. This value is used when partial pressures are in atmospheres (atm).
  
  
  • Caution: Do not use R = 8.314 J mol^-1 K^-1 here unless explicitly stated that pressures are in Pascals and volumes in m³ (which is rare for these problems).
  
  
  
  


• T: Absolute temperature in Kelvin (K). Always convert Celsius to Kelvin (T_K = T_°C + 273.15).


• Δn_g: This is the most critical term. It represents the change in the number of moles of gaseous species during the reaction.
  
  
  
  • Δn_g = (Sum of stoichiometric coefficients of gaseous products) - (Sum of stoichiometric coefficients of gaseous reactants)
  
  
  • Key Point: Only gaseous species are included in the calculation of Δn_g. Solids and liquids are excluded.
  
  
  
  

Impact of Δn_g on Kp and Kc

The value of Δn_g determines the relationship between Kp and Kc:

• 
  Case 1: Δn_g = 0
  
  
  
  • If the number of moles of gaseous products equals the number of moles of gaseous reactants, then (RT)⁰ = 1.
  
  
  • In this case, Kp = Kc.
  
  
  • Example: H₂(g) + I₂(g) ⇌ 2HI(g) (Δn_g = 2 - (1+1) = 0)
  
  
  
  


• 
  Case 2: Δn_g > 0
  
  
  
  • If the number of moles of gaseous products is greater than that of gaseous reactants, then (RT)^Δn_g > 1 (at T > 0 K).
  
  
  • In this case, Kp > Kc.
  
  
  • Example: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) (Δn_g = (1+1) - 1 = 1)
  
  
  
  


• 
  Case 3: Δn_g < 0
  
  
  
  • If the number of moles of gaseous products is less than that of gaseous reactants, then (RT)^Δn_g < 1 (at T > 0 K).
  
  
  • In this case, Kp < Kc.
  
  
  • Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (Δn_g = 2 - (1+3) = -2)
  
  
  
  

Exam Relevance (CBSE & JEE Main)

• CBSE: Expect questions that require you to state the relationship, define terms, and apply the formula to simple reactions. Derivation might also be asked.


• JEE Main: Focus will be on applying the formula to calculate Kp from Kc (or vice versa), often involving multi-step calculations or choosing the correct value of R and Δn_g. Miscalculating Δn_g or using the wrong R value are common pitfalls.

Mastering this formula and the correct calculation of Δn_g is crucial for scoring well in equilibrium problems.

Problem Solving Approach for $K_p$ and $K_c$ Relationship

Understanding the relationship between the equilibrium constants $K_p$ (for pressures) and $K_c$ (for concentrations) is crucial for solving problems in chemical equilibrium. The fundamental equation connecting them is:

$K_p = K_c (RT)^{Delta n_g}$

Where:
* $K_p$: Equilibrium constant expressed in terms of partial pressures.
* $K_c$: Equilibrium constant expressed in terms of molar concentrations.
* R: Ideal gas constant.
* T: Absolute temperature in Kelvin.
* $Delta n_g$: Change in the number of moles of gaseous products and reactants.

Here's a step-by-step approach to solve problems involving $K_p$ and $K_c$:

1. Write a Balanced Chemical Equation:
   
   
   
   • Ensure the given chemical reaction is balanced. This is the first and most critical step.
   
   
   
   


2. Calculate $Delta n_g$:
   
   
   
   • $Delta n_g = ( ext{Sum of stoichiometric coefficients of gaseous products}) - ( ext{Sum of stoichiometric coefficients of gaseous reactants})$.
   
   
   • Crucial Point: Only consider gaseous reactants and products when calculating $Delta n_g$. Solids, liquids, and aqueous species are ignored.
   
   
   • For example, for the reaction $N_2(g) + 3H_2(g)
     ightleftharpoons 2NH_3(g)$:
     
     
     
     • Gaseous products: $2NH_3$ (coefficient = 2)
     
     
     • Gaseous reactants: $1N_2 + 3H_2$ (coefficients = 1 + 3 = 4)
     
     
     • $Delta n_g = 2 - 4 = -2$.
     
     
     
     
   
   
   
   


3. Determine the Value of R (Ideal Gas Constant):
   
   
   
   • The choice of R depends on the units of pressure used in the problem (implicitly or explicitly for $K_p$) and the desired units of $K_p$ or $K_c$.
   
   
   • If pressure is in atmospheres (atm): $R = 0.0821 ext{ L atm mol}^{-1} ext{ K}^{-1}$.
   
   
   • If pressure is in Pascals (Pa) or kilopascals (kPa), and volume in $m^3$: $R = 8.314 ext{ J mol}^{-1} ext{ K}^{-1}$ (or $ ext{Pa m}^3 ext{ mol}^{-1} ext{ K}^{-1}$).
   
   
   • Common Mistake: Using the wrong R value can lead to incorrect results. Always ensure unit consistency.
   
   
   
   


4. Convert Temperature to Kelvin (T):
   
   
   
   • Temperature must always be in Kelvin. If given in Celsius, convert using the formula: $T(K) = T(^circ C) + 273.15$.
   
   
   
   


5. Substitute and Solve:
   
   
   
   • Once all values ($K_c$ or $K_p$, R, T, $Delta n_g$) are known, substitute them into the equation $K_p = K_c (RT)^{Delta n_g}$ and solve for the unknown equilibrium constant.
   
   
   
   

JEE & CBSE Focus:

• This relationship is fundamental for both JEE Main/Advanced and CBSE board exams. Questions often test the calculation of $Delta n_g$ and the correct application of R and T.


• Tip for JEE: Sometimes, the question might provide $K_p/K_c$ at one temperature and ask about it at another, requiring an understanding of temperature's role. Also, be quick in identifying gaseous species.

Example:

Consider the reaction: $2SO_2(g) + O_2(g)
ightleftharpoons 2SO_3(g)$ at $227^circ C$. If $K_c = 100 ext{ M}^{-1}$, find $K_p$.

Step	Action	Calculation / Result
1. Balanced Equation	$2SO_2(g) + O_2(g) ightleftharpoons 2SO_3(g)$	Already balanced.
2. Calculate $Delta n_g$	(Gaseous products) - (Gaseous reactants)	$Delta n_g = (2) - (2+1) = 2 - 3 = -1$.
3. Convert T to Kelvin	$T(K) = T(^circ C) + 273.15$	$T = 227 + 273.15 = 500.15 ext{ K} approx 500 ext{ K}$.
4. Choose R value	Since $K_c$ is in Molarity ($mol/L$) and we'll express pressure in atm.	$R = 0.0821 ext{ L atm mol}^{-1} ext{ K}^{-1}$.
5. Substitute and Solve	$K_p = K_c (RT)^{Delta n_g}$	$K_p = 100 imes (0.0821 imes 500)^{-1}$ $K_p = 100 imes (41.05)^{-1}$ $K_p = 100 / 41.05 approx 2.436 ext{ atm}^{-1}$

Keep practicing to master this essential relationship!

CBSE Focus Areas: Relationship between Kp and Kc

For CBSE Board examinations, the relationship between Kp and Kc is a crucial topic, often tested both conceptually and through direct derivations or numerical problems. Mastering this section requires a clear understanding of the formula, its derivation, and the significance of each term.

1. The Fundamental Relationship

The quantitative relationship between the equilibrium constants expressed in terms of partial pressures (Kp) and molar concentrations (Kc) for a gaseous reversible reaction is given by:

Kp = Kc (RT)^Δn_g

Where:

• Kp: Equilibrium constant in terms of partial pressures.


• Kc: Equilibrium constant in terms of molar concentrations.


• R: Ideal Gas Constant (0.0821 L atm mol^-1 K^-1 or 8.314 J mol^-1 K^-1). Choose R based on units of pressure/volume.


• T: Absolute temperature in Kelvin.


• Δn_g: Change in the number of moles of gaseous products minus the number of moles of gaseous reactants.

2. Key CBSE Focus: Derivation of the Relationship

A common and important question in CBSE exams is the derivation of Kp = Kc (RT)^Δn_g. This requires starting from the ideal gas equation and substituting concentration terms.

Consider a general reversible gaseous reaction:
aA(g) + bB(g) ⇌ cC(g) + dD(g)

1. Expression for Kc:

Kc = [C]^c [D]^d[A]^a [B]^b

2. Expression for Kp:

Kp = (P_C)^c (P_D)^d(P_A)^a (P_B)^b

3. Using Ideal Gas Equation: For ideal gases, PV = nRT, which implies P = nV RT. Since nV is molar concentration [C], we have P = [C]RT.
So, for each gaseous component:

• P_A = [A]RT


• P_B = [B]RT


• P_C = [C]RT


• P_D = [D]RT

4. Substituting into Kp expression:

Kp = ([C]RT)^c ([D]RT)^d([A]RT)^a ([B]RT)^b

Kp = [C]^c [D]^d[A]^a [B]^b × (RT)^c (RT)^d(RT)^a (RT)^b

Kp = [C]^c [D]^d[A]^a [B]^b × (RT)^{(c+d) - (a+b)}

5. Final step: Recognizing the Kc expression and defining Δn_g:

Kp = Kc (RT)^Δn_g

Where Δn_g = (moles of gaseous products) - (moles of gaseous reactants) = (c+d) - (a+b).

3. Significance of Δn_g for CBSE

The value of Δn_g dictates the relationship between Kp and Kc:

• If Δn_g = 0: Kp = Kc (RT)⁰ = Kc. This occurs when the total number of moles of gaseous products equals the total number of moles of gaseous reactants (e.g., H₂(g) + I₂(g) ⇌ 2HI(g)).


• If Δn_g > 0: Kp > Kc. This occurs when the total number of moles of gaseous products is greater than that of gaseous reactants (e.g., PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Δn_g = 1).


• If Δn_g < 0: Kp < Kc. This occurs when the total number of moles of gaseous products is less than that of gaseous reactants (e.g., N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Δn_g = -2).

CBSE Tip: Always pay attention to the phases of reactants and products. Only gaseous species contribute to Δn_g.

4. Numerical Problem Solving

CBSE often includes numerical problems where you are given one equilibrium constant (Kp or Kc), the temperature, and the balanced chemical equation, and asked to calculate the other.

Ensure you use the correct value of R and convert temperature to Kelvin.

For example, if pressures are in atm, use R = 0.0821 L atm mol^-1 K^-1.

Practice is key for this section to score well in the board exams!

Relationship between Kp and Kc: JEE Focus Areas

Understanding the relationship between Kp and Kc is crucial for JEE Main and Advanced, as it frequently appears in both direct formula-based questions and as a part of larger equilibrium problems. Mastering this concept requires precision, especially in calculating the change in the number of moles of gaseous species.

1. The Core Relationship

The fundamental equation connecting Kp (equilibrium constant in terms of partial pressures) and Kc (equilibrium constant in terms of molar concentrations) is:

Kp = Kc (RT)^Δng

• Kp: Equilibrium constant expressed in terms of partial pressures of gaseous reactants and products.


• Kc: Equilibrium constant expressed in terms of molar concentrations of reactants and products.


• R: Ideal gas constant.
  
  JEE Tip: For Kp-Kc conversions, always use R = 0.0821 L atm mol⁻¹ K⁻¹. Using 8.314 J mol⁻¹ K⁻¹ (the SI unit) will lead to incorrect results unless energy terms are involved elsewhere, which is rare for this specific conversion.
  


• T: Temperature in Kelvin (K). Always convert Celsius to Kelvin (T_K = T_°C + 273.15).


• Δng: Change in the number of moles of gaseous products minus the number of moles of gaseous reactants. This is the most critical term for JEE.

2. Calculating Δng: The Most Common Pitfall

The correct calculation of Δng is paramount.

Δng = (Sum of stoichiometric coefficients of gaseous products) - (Sum of stoichiometric coefficients of gaseous reactants)

Key Considerations:

• Only Gaseous Species: Solids and liquids are excluded from the calculation of Δng. This is a frequent error point in JEE. For example, in the reaction:
  
  CaCO₃(s) ↔ CaO(s) + CO₂(g)
  
  Δng = (moles of CO₂(g)) - (moles of gaseous reactants) = 1 - 0 = 1.
  


• Stoichiometric Coefficients: Use the coefficients from the balanced chemical equation.

Cases for Δng:

• If Δng = 0: Kp = Kc (RT)⁰ ⇒ Kp = Kc. This occurs when the total moles of gaseous products equal the total moles of gaseous reactants.
  
  Example: H₂(g) + I₂(g) ↔ 2HI(g); Δng = 2 - (1+1) = 0.
  


• If Δng > 0: Kp = Kc (RT)^{positive value}. Since RT is generally > 1 at common temperatures, Kp > Kc.
  
  Example: N₂O₄(g) ↔ 2NO₂(g); Δng = 2 - 1 = 1.
  


• If Δng < 0: Kp = Kc (RT)^{negative value} ⇒ Kp = Kc / (RT)^|Δng|. In this case, Kp < Kc.
  
  Example: N₂(g) + 3H₂(g) ↔ 2NH₃(g); Δng = 2 - (1+3) = -2.
  

3. JEE Specific Problem-Solving Strategy

1. Balance the Equation: Ensure the given chemical equation is balanced.


2. Identify Gaseous Species: Clearly mark all gaseous reactants and products.


3. Calculate Δng: Sum coefficients of gaseous products and subtract the sum of coefficients of gaseous reactants. Be meticulous.


4. Convert Temperature to Kelvin: T(K) = T(°C) + 273.15.


5. Use Correct R Value: R = 0.0821 L atm mol⁻¹ K⁻¹.


6. Substitute and Solve: Plug values into Kp = Kc (RT)^Δng to find the unknown constant.

JEE Focus: Questions often test your ability to correctly calculate Δng for various heterogeneous and homogeneous equilibria. Pay close attention to the physical states of all species.

Keep practicing problems involving all three cases of Δng to build confidence and accuracy. Good luck!