Chapter 18, Problem 59GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Methanol is now widely used as a fuel in race cars. Consider the following reaction as a possible synthetic route to methanol.C(graphite) + ½ O2(g) + 2 H2(g) ⇄ CH3OH(ℓ)Calculate Kp for the formation of methanol at 298 K using this reaction. Would this reaction be more product-favored at a different temperature?

Interpretation Introduction

Interpretation:

The Kp for methanol formation should be calculated and should be identified that whether the reaction will be more product favoured at different temperatures.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrG. It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is:

ΔrG°=nΔfG°(products)nΔfG°(reactants)

ΔrG is related to the equilibrium constant K by the equation,

ΔrG=RTlnKp

The rearranged expression is,

Kp=eΔrGRT

Explanation

The ΔrG° and the equilibrium constant for the given reaction is calculated below.

The Appendix L referred for the values of standard free energy values.

The ΔfG° for C(graphite) is 0 kJ/mol.

The ΔfG° for O2(g) is 0 kJ/mol.

The ΔfG° for H2(g) is 0 kJ/mol.

The ΔfG° for CH3OH(l) is 166.14 kJ/mol.

The given reaction is,

C(graphite)+12O2(g)+2H2(g)CH3OH(l)

ΔrG°=nΔfG°(products)nΔfG°(reactants)=[(1 mol CH3OH(l)/mol-rxn)ΔfG°[CH3OH(l)][(1 mol C(graphite)/mol-rxn)ΔfG°[C(graphite)]+(0

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