   # Δ H f ° for iodine gas is 62.4 kJ/mol, and S ° is 260.7 J/mol ⋅ K. Calculate the equilibrium partial pressures of I 2 ( g ), H 2 ( g ), and HI(g) for the system 2 HI ( g ) ⇌ H 2 ( g ) + I 2 ( g ) at 500°C if the initial partial pressures are all 0.200 atm. ### Chemistry: Principles and Reactions

8th Edition
William L. Masterton + 1 other
Publisher: Cengage Learning
ISBN: 9781305079373

#### Solutions

Chapter
Section ### Chemistry: Principles and Reactions

8th Edition
William L. Masterton + 1 other
Publisher: Cengage Learning
ISBN: 9781305079373
Chapter 16, Problem 94QAP
Textbook Problem
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## Δ H f ° for iodine gas is 62.4 kJ/mol, and S° is 260.7 J/mol ⋅ K. Calculate the equilibrium partial pressures of I2(g), H2(g), and HI(g) for the system 2 HI ( g ) ⇌ H 2 ( g ) + I 2 ( g ) at 500°C if the initial partial pressures are all 0.200 atm.

Interpretation Introduction

Interpretation:

For the given reaction,theequilibrium partial pressures of I2, H2 and HI are to be determined at T = 500 °C and with 0.200 atm asthe initial pressure of all species.

2HI(g) H2(g) + I2(g)

Concept introduction:

1. The equilibrium constant (Keq) for a given reaction is a parameter which is expressed as a ratio of the concentration (or pressure in the case of gases) of products to that of the reactants.
2. The equilibrium constant is related to the standard Gibbs free energyΔG° by the following equation:
3. ΔG0 = -RTlnKeq -----(1)

4. The temperature dependence of Keq can be expressed in terms of the Von’t Hoff equation:
5. lnK2K1 = ΔHR[1T1-1T2] ------(2)

### Explanation of Solution

Given Information:

ΔHf0(I2) = 62.4 kJ/molS0(I2) = 260.7 J/K-mol

Calculation:

Step 1: CalculateΔG°

The given reaction is:

2HI(g) H2(g) + I2(g)

The standard enthalpy change for a reaction ΔH0 is given in terms of the difference in the standard enthalpy of formation of the products and that of reactants.

ΔH0 = npΔHf0(products) - nrΔHf0(reactants)

Where np and nr are the number of moles of the products and reactants

For the above reaction:

ΔH0 = [1ΔHf0(H2(g)) + 1ΔHf0(I2(g))] - [2ΔHf0(HI(g)) ]Based on the ΔHf0 values:ΔH0 = [1 mole ×(0 kJ/mol) + 1 mole ×(62.4 kJ/mol)]-[2 mole ×(26.48 kJ/mol)]         = 0 +62.4 - 52.96  = 9.44 kJ

The standard entropy change for a reaction ΔS0 is given in terms of the difference in the standard entropy of formation of the products and that of reactants.

ΔS0 = npSf0(products) - nrSf0(reactants)

Where np and nr are the number of moles of the products and reactants

For the above reaction:

ΔS0 = [Sf0(H2(g)) + 1Sf0(I2(g))] - [2Sf0(HI(g)) ]Based on the Sf0 values:ΔS0 = [1 mole ×(130.57 J/mol-K) + 1 mole ×(260.7 J/mol-K)]-[2 mole ×(206.48 J/mol-K)]         = 130.57 + 260.7 - 412.96  = -21.69 J/mol-K

At T = 298 K,

ΔG0 = ΔH0 - TΔS0=9.44(298)(0.02169)=15.90 kJ

Step 2: Calculate Keq at 298 K

ΔG0 = -RTlnKK = e-ΔG0RTΔG0 =15

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