   Chapter 18, Problem 47PS

Chapter
Section
Textbook Problem

Calculate ΔrG° at 25 °C for the formation of 1.00 mol of C2H6(g) from C2H4(g) and H2(g). Use this value to calculate Kp for the equilibrium.C2H4(g) + H2(g) ⇄ C2H6(g)Comment on the sign of ΔrG° and the magnitude of Kp.

Interpretation Introduction

Interpretation:

The ΔrG° 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) + H2(g)2C2H6(g)

The ΔrG° for C2H6(g) is 31.89 kJ/mol.

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

The ΔrG° for H2(g) is 0 kJ/mol.

ΔrG°=nΔfG°(products)nΔfG°(reactants)=[(2 mol C2H6(g)/mol-rxn)ΔfG°[C2H6(g)][(1 mol C2H4(g)/mol-rxn)ΔfG°[C2H4(g)]+(1 mol H2(g)/mol-rxn)

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

Find more solutions based on key concepts 