Chapter 12, Problem 23QRT

(a)

Interpretation Introduction

Interpretation:

The equilibrium constant expression for the reaction $2{\text{SO}}_{\text{3(g)}}\rightleftharpoons {\text{ 2SO}}_{\text{2(g)}}{\text{ + O}}_{\text{2(g)}}$ has to be written and it has to be named as ${\text{K}}_{\text{1}}$.

Concept Introduction:

Equilibrium constant: At equilibrium the ratio of products to reactants (each raised to the power corresponding to its stoichiometric coefficient) has a constant value, K

For a general reaction,

  $\text{aA+bB}\rightleftarrows \text{cC+dD}$

  $\text{Equilibrium}\text{constant}\text{K}\text{=}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$

The concentration of solids and pure liquids do not change, so their concentration terms are not included in the equilibrium constant expression.

(b)

Interpretation Introduction

Interpretation:

The equilibrium constant expression for the decomposition of one mole of ${\text{SO}}_{\text{3}}$ to ${\text{SO}}_2$ and ${\text{O}}_2$ has to be written and it has to be named as ${\text{K}}_2$.

Concept Introduction:

Refer part (a) for concept.

(c)

Interpretation Introduction

Interpretation:

The relation between ${\text{K}}_{\text{1}}$ and ${\text{K}}_2$ has to be given.

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.