Calculate the equilibrium constant for the formation of NiO at 1627 °C. Can the reaction proceed in the forward direction if the initial pressure of O 2 is below 1.00 mm Hg? {Δ f G° [NiO(s)] = −72.1 kJ/mol at 1627 °C} Ni(s) + ½ O 2 (g) ⇄ NiO(s)

Chemistry & Chemical Reactivity

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

Chapter
Section

Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 18, Problem 64IL
Textbook Problem
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Calculate the equilibrium constant for the formation of NiO at 1627 °C. Can the reaction proceed in the forward direction if the initial pressure of O2 is below 1.00 mm Hg? {ΔfG° [NiO(s)] = −72.1 kJ/mol at 1627 °C}Ni(s) + ½ O2(g) ⇄ NiO(s)

Interpretation Introduction

Interpretation:

The equilibrium constant for the formation of NiO at 1627 °C should be calcualted under given conditions.

Concept introduction:

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

ΔrG°fG°(products)fG°(reactants)

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

ΔfGo= -RTlnKp

The rearranged expression is,

Kp= e-ΔfGoRT

ΔfGo is also related to the reaction quotient Q by the expression,

ΔfG = ΔfG°+RTlnQ

For a general reaction, aA + bBcC + dD

Q = [C]c[D]d[A]a[B]b

Explanation of Solution

The equilibrium constant for the formation of NiO at 1627 °C is calculated below.

Given:

The value of ΔfGo for NiO(s) is 72.1 kJ/mol-rxn

ΔGo is related to the equilibrium constant K by the equation, ΔfGo= -RTlnKp

The rearranged expression is, Kp= e-ΔfGoRT

Substituting the values of ΔfG°, T and R into above rearranged equation as follows,

Kp= e--72.1 kJ/mol-rxn(0.008314 kJ/K×mol)(1900.15 K)= 90

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