8.3.1 Equilibrium Expressions Involving Partial Pressure, Number of Moles and Mole Fraction

This topic explains how equilibrium expressions can be written in terms of partial pressure, number of moles, and mole fraction. It covers the relationship between K_p and K_c.

Definition

Partial Pressure: The pressure exerted by an individual gas in a mixture of gases.

Mole Fraction: The ratio of the number of moles of a component to the total number of moles of all components in a mixture.

K_p: The equilibrium constant expressed in terms of partial pressures.

K_c: The equilibrium constant expressed in terms of molar concentrations.

Main Idea

For gaseous reversible reactions, the equilibrium expression is often expressed in terms of partial pressure of each gas. According to Henry's law, at constant temperature, the partial pressure of a gas is directly proportional to its molar concentration. The relationship between K_p and K_c is given by:

K_p = K_c (RT)^{Δ n}

Where Δ n = (moles of gaseous products) − (moles of gaseous reactants).

Core Concept

• K_p: Equilibrium constant in terms of partial pressures.
• K_c: Equilibrium constant in terms of molar concentrations.
• Relationship: K_p = K_c (RT)^{Δ n}
• Δ n: Change in number of moles of gas.
• R: Gas constant = 0.0821 dm³ atm K⁻¹ mol⁻¹
• T: Temperature in Kelvin.

Detailed Explanation

Equilibrium Expressions Involving Partial Pressure

Consider the general gaseous reversible reaction:

aA_(g) + bB_(g) ⇌ cC_(g) + dD_(g)

For gases, the expression is often expressed in terms of partial pressure of each gas. According to Henry's law, "At constant temperature, the partial pressure of a gas is directly proportional to its molar concentration."

P ∝ C

P = CRT

Where R is the gas constant.

Relationship Between K_p and K_c

K_p = K_c (RT)^{Δ n}

• Δ n = (c + d) - (a + b) = change in number of moles of gas
• R = 0.0821 dm³ atm K^-1 mol^-1
• T = temperature in Kelvin

Steps for Solving Problems Involving K_p and K_c

• Write K_p expression.
• Substitute the partial pressures of each species.
• Simplify to get K_p.
• Calculate Δ n.
• Write expression relating K_p and K_c.
• Substitute known values in it and find K_c.

Example 8.4: Calculating K_p and K_c

Problem: For the reaction: 2NO_(g) + Cl_2(g) ⇌ 2NOCl_(g)

• P_NOCl = 1.2 atm
• P_NO = 5.0 × 10^-2 atm
• P_Cl2 = 3.0 × 10^-1 atm

Calculate K_p and K_c at 25°C.

Solution:

Step 1: Write K_p expression K_p = (P_NOCl)²(P_NO)²(P_Cl2)

Step 2: Substitute partial pressures K_p = (1.2)²(5.0 × 10^-2)²(3.0 × 10^-1)

Step 3: Simplify K_p = 1.9 × 10³

Step 4: Calculate Δ n Δ n = 2 - (2 + 1) = -1

Step 5: Write expression relating K_p and K_c K_p = K_c (RT)^{Δ n}

Step 6: Substitute known values R = 0.0821 dm³ atm K^-1 mol^-1 T = 25^° C + 273 = 298 K K_p = K_c (RT)^{Δ n} 1.9 × 10³ = K_c (0.0821 × 298)^-1 1.9 × 10³ = K_c(0.0821 × 298) K_c = 1.9 × 10³ × 0.0821 × 298 K_c = 4.65 × 10⁴

Final Answer: K_p = 1.9 × 10³ K_c = 4.65 × 10⁴

Concept Assessment Exercise 8.2

The contact process prepares purest sulphuric acid commercially. Following reaction takes place in the contact chamber in the presence of V₂O₅ at 450°C.

2SO_2(g) + O_2(g) ⇌ 2SO_3(g)

• [SO₂] = 0.59 M
• [O₂] = 0.05 M
• [SO₃] = 0.259 M

(Ans: 0.0649 atm)

Intermediate Level

K_p \) and \( K_c

Relationship

K_p = K_c (RT)^{Δ n}

Steps for Calculations

• Write K_p expression.
• Substitute partial pressures.
• Simplify to get K_p.
• Calculate Δ n.
• Use K_p = K_c (RT)^{Δ n}.
• Solve for K_c.

Example 8.4

• P_NOCl = 1.2 atm
• P_NO = 5.0 × 10^-2 atm
• P_Cl2 = 3.0 × 10^-1 atm
• T = 25^° C = 298 K

Solution: K_p = (1.2)²(5.0 × 10^-2)²(3.0 × 10^-1) = 1.9 × 10³ Δ n = 2 - 3 = -1 K_c = K_p (RT)^{-Δ n} = 1.9 × 10³ × 0.0821 × 298 = 4.65 × 10⁴

Important Points

• 1. K_p: Equilibrium constant in partial pressures.
• 2. K_c: Equilibrium constant in concentrations.
• 3. K_p = K_c (RT)^{Δ n}
• 4. Δ n = moles gaseous products − moles gaseous reactants.
• 5. R = 0.0821 dm³ atm K^-1 mol^-1
• 6. Example 8.4: K_p = 1.9 × 10³, K_c = 4.65 × 10⁴.
• 7. Concept Assessment 8.2: K_p = 0.0649 atm.

Remember

• - K_p: Partial pressures
• - K_c: Concentrations
• - K_p = K_c (RT)^{Δ n}
• - Δ n: Change in moles of gas
• - Example 8.4: K_p = 1.9 × 10³, K_c = 4.65 × 10⁴

Common Mistakes

• 1. Forgetting to calculate Δ n: Essential for relating K_p and K_c.
• 2. Mixing up K_p and K_c: K_p uses pressures; K_c uses concentrations.
• 3. Forgetting units of R: R = 0.0821 dm³ atm K^-1 mol^-1
• 4. Miscalculating Δ n: Products − reactants.

Quick Revision

• K_p: Equilibrium constant in partial pressures.
• K_c: Equilibrium constant in concentrations.
• K_p = K_c (RT)^{Δ n}
• Δ n = moles gaseous products − moles gaseous reactants.
• R = 0.0821 dm³ atm K^-1 mol^-1
• Example 8.4: K_p = 1.9 × 10³, K_c = 4.65 × 10⁴.
• Concept Assessment 8.2: K_p = 0.0649 atm.

60-Second Revision

• K_p: Partial pressures
• K_c: Concentrations
• K_p = K_c (RT)^{Δ n}
• Δ n: Change in moles of gas
• Example 8.4: K_p = 1.9 × 10³, K_c = 4.65 × 10⁴

QUALITY VALIDATION

• Topic: 8.3.1 Equilibrium Expressions Involving Partial Pressure, Number of Moles and Mole Fraction
• Chapter: 8 (Chemical Equilibrium)

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• Derivations checked: PASS
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