Chapter 13, Problem 13.75SP

(a)

Interpretation Introduction

Interpretation:

The reaction quotient of given reaction has to be calculated. Also the direction of reaction to reach equilibrium has to be predicted.

Concept introduction:

Reaction quotient (Q):

Reaction quotient (Q) is the ratio of product of concentration of products to that of product of concentration of reactants, raised to the power of coefficients.

Consider a reaction

$\text{aA + bB}\stackrel{}{\rightleftharpoons }\text{cC + dD}$

$\text{Reaction}\text{quotient}{\text{Q}}_{\text{c}}\text{=}\frac{{\text{[C]}}_{\text{t}}{}^{\text{c}}{\text{[D]}}_{\text{t}}{}^{\text{d}}}{{\text{[A]}}_{\text{t}}{}^{\text{a}}{\text{[B]}}^{\text{b}}}$

Where,

a, b, c and d are coefficients of each species.

t is arbitrary time for measuring concentration .

Figure 1

(b)

Interpretation Introduction

Interpretation:

At equilibrium the molar concentrations of ${\text{BrCl, Br}}_{\text{2}}\text{ and}{\text{Cl}}_{\text{2}}$ has to be calculated.

Concept introduction:

Calculation of equilibrium concentration from initial concentration:

• The balanced equation for given reaction has to be written.
• Set up an equilibrium table to calculate x value
• The value of x is solved by substituting concentrations of chemical species in equilibrium constant expression.
• The concentration of chemical species at equilibrium is calculated from value of x

Equilibrium constant$\left({\text{K}}_{\text{c}}\right)$:

Equilibrium constant $\left({\text{K}}_{\text{c}}\right)$ is the ratio of the rate constants of the forward and reverse reactions at a given temperature. In other words it is the ratio of the concentrations of the products to concentrations of the reactants. Each concentration term is raised to a power, which is same as the coefficients in the chemical reaction.

Consider the reaction where A reacts to give B.

$\text{aA}\rightleftharpoons \text{bB}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}{\left[\text{A}\right]}^{\text{a}}{\text{=k}}_{\text{r}}{\left[\text{B}\right]}^{\text{b}}\end{array}$

On rearranging,

$\frac{{\left[\text{B}\right]}^{\text{b}}}{{\left[\text{A}\right]}^{\text{a}}}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{c}}$

Where,

${\text{k}}_{\text{f}}$ is the rate constant of the forward reaction.

${\text{k}}_{\text{r}}$ is the rate constant of the reverse reaction.

${\text{K}}_{\text{c}}$ is the equilibrium constant.