   Chapter 18, Problem 82IL

Chapter
Section
Textbook Problem

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

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

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 