Chapter 17.2, Problem 17.2BFP

(a)

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{c}}$ value for the reaction of $4{\text{NOBr}}_{\text{(g)}}\underset{}{\rightleftharpoons }{\text{ 4NO}}_{\text{(g)}}{\text{ + 2Br}}_{\text{2(g)}}$ 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.

(b)

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{c}}$ value for the reaction of $\frac{1}{2}{\text{NO}}_{\text{(g)}}\text{ + }\frac{1}{4}{\text{Br}}_{\text{2(g)}}\underset{}{\rightleftharpoons }\frac{1}{2}{\text{NOBr}}_{\text{(g)}}$ 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.