General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
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Chapter 23, Problem 23.35P

(a)

Interpretation Introduction

Interpretation:

Value of the equilibrium constant and ΔGrxno has to be determined for the reaction CO(g) + 2H2(g)CH3OH(l).

Concept Introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔG.

Value of ΔGrxno can be calculated by the equation:

  ΔGrxnoG°[products]G°[reactants]

Equilibrium constant of a reaction can be determined from the value of ΔGrxno using the given formula,

  ΔGrxno=RTln K

(a)

Expert Solution
Check Mark

Answer to Problem 23.35P

Value of ΔGrxno is 29.4 kJ/mol.

Value of the equilibrium constant is 142.3×103.

Explanation of Solution

The given reaction is shown below,

  CO(g) + 2H2(g)CH3OH(l)

  • Calculate the value of ΔGrxno:

Standard free energy of formation values is given below,

  ΔG°[CO(g)]=137.2 kJ/molΔG°[H2(g)]=0 kJ/molΔG°[CH3OH(l)]=166.6 kJ/mol

Value of ΔGrxno can be calculated by the equation:

  ΔGrxnoG°[products]G°[reactants]

Substitute the values as follows,

  ΔGrxno[(1)(166.6 kJ/mol)][(1)(137.2 kJ/mol)+(2)(0 kJ/mol)]=29.4 kJ/mol

Therefore, the value of ΔGrxno is 29.4 kJ/mol.

  • Calculate the value of the equilibrium constant:

Value of the equilibrium constant is determined as follows,

  T = 25°C(25+273) = 298 KΔGrxno=RTln K29.4 kJ/mol =(8.314×103 kJ.K1.mol1)(298 K)ln Kln K=29.4 kJ/mol(8.314×103 kJ.K1.mol1)(298 K) = 11.866K = 142.3×103

Therefore, the value of the equilibrium constant is 142.3×103.

(b)

Interpretation Introduction

Interpretation:

Value of the equilibrium constant and ΔGrxno has to be determined for the reaction C(s, graphite) + H2O(g)CO(g)+ H2(g).

Concept Introduction:

Refer to (a).

(b)

Expert Solution
Check Mark

Answer to Problem 23.35P

Value of ΔGrxno is 91.4 kJ/mol.

Value of the equilibrium constant is 9.51×1017.

Explanation of Solution

The given reaction is shown below,

  C(s, graphite) + H2O(g)CO(g)+ H2(g)

  • Calculate the value of ΔGrxno:

Standard free energy of formation values is given below,

  ΔG°[C(s, graphite)]=0 kJ/molΔG°[H2O(g)]=228.6 kJ/molΔG°[CO(g)]=137.2 kJ/molΔG°[H2(g)]=0 kJ/mol

Value of ΔGrxno can be calculated by the equation:

  ΔGrxnoG°[products]G°[reactants]

Substitute the values as follows,

  ΔGrxno[(1)(137.2 kJ/mol)+(1)(0 kJ/mol)] [(1)(0 kJ/mol)+(1)(228.6 kJ/mol)]=91.4 kJ/mol

Therefore, the value of ΔGrxno is 91.4 kJ/mol.

  • Calculate the value of the equilibrium constant:

Value of the equilibrium constant is determined as follows,

  T = 25°C(25+273) = 298 KΔGrxno=RTln K91.4 kJ/mol =(8.314×103 kJ.K1.mol1)(298 K)ln Kln K=91.4 kJ/mol(8.314×103 kJ.K1.mol1)(298 K) = 36.891K = 9.51×1017

Therefore, the value of the equilibrium constant is 9.51×1017.

(c)

Interpretation Introduction

Interpretation:

Value of the equilibrium constant and ΔGrxno has to be determined for the reaction CO(g) + 2H2(g)CH4(g) + H2O(g).

Concept Introduction:

Refer to (a).

(c)

Expert Solution
Check Mark

Answer to Problem 23.35P

Value of ΔGrxno is 141.9 kJ/mol.

Value of the equilibrium constant is 7.48×1028.

Explanation of Solution

The given reaction is shown below,

  CO(g) + 2H2(g)CH4(g) + H2O(g)

  • Calculate the value of ΔGrxno:

Standard free energy of formation values is given below,

  ΔG°[CO(g)]=137.2 kJ/molΔG°[H2(g)]=0 kJ/molΔG°[CH4(g)]=50.5 kJ/molΔG°[H2O(g)]=228.6 kJ/mol

Value of ΔGrxno can be calculated by the equation:

  ΔGrxnoG°[products]G°[reactants]

Substitute the values as follows,

  ΔGrxno[(1)(50.5 kJ/mol)+(1)(228.6 kJ/mol)] [(1)(137.2 kJ/mol)+(3)(0 kJ/mol)]=141.9 kJ/mol

Therefore, the value of ΔGrxno is 141.9 kJ/mol.

  • Calculate the value of the equilibrium constant:

Value of the equilibrium constant is determined as follows,

  T = 25°C(25+273) = 298 KΔGrxno=RTln K141.9 kJ/mol =(8.314×103 kJ.K1.mol1)(298 K)ln Kln K=141.9 kJ/mol (8.314×103 kJ.K1.mol1)(298 K) = 57.2738K = 7.48×1028

Therefore, the value of the equilibrium constant is 7.48×1028.

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Chapter 23 Solutions

General Chemistry

Ch. 23 - Prob. 23.11PCh. 23 - Prob. 23.12PCh. 23 - Prob. 23.13PCh. 23 - Prob. 23.14PCh. 23 - Prob. 23.15PCh. 23 - Prob. 23.16PCh. 23 - Prob. 23.17PCh. 23 - Prob. 23.18PCh. 23 - Prob. 23.19PCh. 23 - Prob. 23.20PCh. 23 - Prob. 23.21PCh. 23 - Prob. 23.22PCh. 23 - Prob. 23.23PCh. 23 - Prob. 23.24PCh. 23 - Prob. 23.25PCh. 23 - Prob. 23.26PCh. 23 - Prob. 23.27PCh. 23 - Prob. 23.28PCh. 23 - Prob. 23.29PCh. 23 - Prob. 23.30PCh. 23 - Prob. 23.31PCh. 23 - Prob. 23.32PCh. 23 - Prob. 23.33PCh. 23 - Prob. 23.34PCh. 23 - Prob. 23.35PCh. 23 - Prob. 23.36PCh. 23 - Prob. 23.37PCh. 23 - Prob. 23.38PCh. 23 - Prob. 23.39PCh. 23 - Prob. 23.40PCh. 23 - Prob. 23.41PCh. 23 - Prob. 23.42PCh. 23 - Prob. 23.43PCh. 23 - Prob. 23.44PCh. 23 - Prob. 23.45PCh. 23 - Prob. 23.46PCh. 23 - Prob. 23.47PCh. 23 - Prob. 23.48PCh. 23 - Prob. 23.49PCh. 23 - Prob. 23.50PCh. 23 - Prob. 23.51PCh. 23 - Prob. 23.52PCh. 23 - Prob. 23.53PCh. 23 - Prob. 23.54PCh. 23 - Prob. 23.55PCh. 23 - Prob. 23.56PCh. 23 - Prob. 23.57PCh. 23 - Prob. 23.58PCh. 23 - Prob. 23.59PCh. 23 - Prob. 23.60PCh. 23 - Prob. 23.61PCh. 23 - Prob. 23.62PCh. 23 - Prob. 23.63PCh. 23 - Prob. 23.64PCh. 23 - Prob. 23.65PCh. 23 - Prob. 23.66PCh. 23 - Prob. 23.67PCh. 23 - Prob. 23.68PCh. 23 - Prob. 23.69PCh. 23 - Prob. 23.70PCh. 23 - Prob. 23.71PCh. 23 - Prob. 23.72PCh. 23 - Prob. 23.73PCh. 23 - Prob. 23.74PCh. 23 - Prob. 23.75PCh. 23 - Prob. 23.76PCh. 23 - Prob. 23.77PCh. 23 - Prob. 23.78PCh. 23 - Prob. 23.79PCh. 23 - Prob. 23.80PCh. 23 - Prob. 23.81PCh. 23 - Prob. 23.82PCh. 23 - Prob. 23.83PCh. 23 - Prob. 23.84PCh. 23 - Prob. 23.85PCh. 23 - Prob. 23.86PCh. 23 - Prob. 23.87PCh. 23 - Prob. 23.88P
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