Chapter 18, Problem 48PS

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

Calculate ΔrG° at 25 °C for the formation of 1.00 mol of C2H5OH(g) from C2H4(g) and H2O(g). Use this value to calculate Kp for the equilibrium.C2H4(g) + H2O(g) ⇄ C2H5OH(g)Comment on the sign of ΔrG° and the magnitude of Kp.

Interpretation Introduction

Interpretation:

The ΔrG° value and the equilibrium constant for the given reaction should be calculated.

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)

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

ΔrG=RTlnKp.

The rearranged expression is,

Kp=eΔrGRT

Explanation

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

Given:

The given reaction is,

C2H4(g) + H2O(g)C2H5OH(g)

The ΔrG° for C2H5OH(g) is 168.49 kJ/mol.

The ΔrG° for C2H4(g) is +68.35 kJ/mol.

The ΔrG° for H2O(g) is 228.59 kJ/mol.

ΔrG°=nΔfG°(products)nΔfG°(reactants)=[(1 mol C2H5OH(g)/mol-rxn)ΔfG°[C2H5OH(g)][(1 mol C2H4(g)/mol-rxn)ΔfG°[C2H4(g)]+(1 mol H2O(g)/mol-rxn)ΔfG°[H2O(g)]] ]

Substituting the values,

ΔrG°=[(1 mol C2H5OH(g)/mol-rxn)(168

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started