Chapter 12, Problem 26QRT

(a)

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{c}}$ for the synthesis of $1\text{ mol}$ sulfur trioxide gas has to be calculated.

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 ${\text{K}}_{\text{c}}$ for the decomposition of of $\text{2 mol}$ sulfur trioxide gas has to be calculated.

Concept Introduction:

Refer part (a)