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Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

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Chapter
Section
BuyFindarrow_forward

Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 18, Problem 30PS
Textbook Problem
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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 of Solution

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

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Chapter 18 Solutions

Chemistry & Chemical Reactivity
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