# The standard free energy change, Δ r G °, for the formation of NO(g) from its elements is + 86.58 kJ/mol-rxn at 25 °C. Calculate K p at this temperature for the equilibrium ½ N 2 (g) + ½ O 2 (g) ⇄ NO(g) Comment on the sign of Δ r G ° and the magnitude of K p .

### 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 27PS
Textbook Problem
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## The standard free energy change, ΔrG°, for the formation of NO(g) from its elements is + 86.58 kJ/mol-rxn at 25 °C. Calculate Kp at this temperature for the equilibrium½ N2(g) + ½ O2(g) ⇄ NO(g)Comment on the sign of ΔrG° and the magnitude of Kp.

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction should be calculated. The sign of ΔrG° and magnitude of Kp should be explained.

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 equilibrium constant for the given reaction of formation of NO(g) is calculated below.

Given:

The given reaction is,

12N2(g)+12O2(g)NO(g)

The ΔrG value for the reaction is 86.58 kJ/mol- rxn at 298.15 K.

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

ΔrG=RTlnKp.

The rearranged expression is,

Kp=eΔrGRT

Substitute the value of ΔrG°, T and R

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