Chapter 17, Problem 17.17P

(a)

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{c}}$ value for the reaction of ${\text{NO}}_{\text{(g)}}{\text{ + H}}_{\text{2(g)}}\underset{}{\rightleftharpoons }\frac{1}{2}{\text{N}}_{\text{2}}{}_{\text{(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(g)}}$ has to be calculated from the given reaction.

Concept Introduction:

${\text{K}}_{\text{c}}$ for forward and reverse reaction:

A equilibrium constant (or reaction quotient) for a forward reaction is the reciprocal of the equilibrium constant (or reaction quotient) for the reverse reaction.

  ${\text{K}}_{\text{c(fwd)}}=\frac{1}{{\text{K}}_{\text{c(rev)}}}{\text{ and Q}}_{\text{c(fwd)}}=\frac{1}{{\text{Q}}_{\text{c(rev)}}}$

When a reaction is multiplied by a factor, the equilibrium constant is raised to a power equal to the factor. When a reaction is reversed, the reciprocal of the equilibrium constant is used as the new equilibrium constant.

(b)

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{c}}$ value for the reaction of ${\text{2N}}_{\text{2}}{}_{\text{(g)}}{\text{ + 4H}}_{\text{2}}{\text{O}}_{\text{(g)}}\stackrel{}{\rightleftharpoons }{\text{ 4NO}}_{\text{(g)}}{\text{ + 4H}}_{\text{2(g)}}$ has to be calculated from the given reaction.

Concept Introduction:

${\text{K}}_{\text{c}}$ for forward and reverse reaction:

A equilibrium constant (or reaction quotient) for a forward reaction is the reciprocal of the equilibrium constant (or reaction quotient) for the reverse reaction.

  ${\text{K}}_{\text{c(fwd)}}=\frac{1}{{\text{K}}_{\text{c(rev)}}}{\text{ and Q}}_{\text{c(fwd)}}=\frac{1}{{\text{Q}}_{\text{c(rev)}}}$

When a reaction is multiplied by a factor, the equilibrium constant is raised to a power equal to the factor. When a reaction is reversed, the reciprocal of the equilibrium constant is used as the new equilibrium constant.