At ${\mathbf{425}}^{\mathbf{0}}\mathbf{C}\mathbf{,}{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{18}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}$for the reaction

role="math" localid="1655096121156" $\mathbf{2}{\mathbf{HBr}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }{\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{Br}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$

In one experiment, 0.20 atm of HBr(g), 0.010 atm of${\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$ , and 0.010 atm of${\mathbf{Br}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$ are introduced into a container. Is the reaction at equilibrium? If not, in which direction will it proceed?

The equilibrium equation representing the decomposition of hydrogen bromide (HBr) is as follows:

$\mathbf{2}{\mathbf{HBr}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }{\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{Br}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$

Express the reaction quotient of the reaction as follows:

${\mathrm{Q}}_{\mathrm{p}}=\frac{\left({\mathrm{P}}_{{\mathrm{H}}_2}\right)\left({\mathrm{P}}_{{\mathrm{Br}}_2}\right)}{{\left({\mathrm{P}}_{\mathrm{HBr}}\right)}^2}$

Here, ${\mathrm{P}}_{{\mathrm{H}}_2}$ is the initial partial pressure of ${\mathrm{H}}_2$, ${\mathrm{P}}_{{\mathrm{Br}}_2}$ is the initial partial pressure of${\mathrm{Br}}_2$ , ${\mathrm{P}}_{\mathrm{HBr}}$is the initial partial pressure of$\mathrm{HBr}$ , ${\mathrm{Q}}_{\mathrm{p}}$and is he reaction quotient of the reaction.

Comparison of reaction quotient ${\mathrm{Q}}_{\mathrm{p}}$and the equilibrium constant ${\mathrm{K}}_{\mathrm{p}}$of the reaction.

When${\mathrm{Q}}_{\mathrm{p}}>{\mathrm{K}}_{\mathrm{p}}$ , the reaction proceeds towards the left side, favouring the reactants until the equilibrium is attained.

When, ${\mathrm{Q}}_{\mathrm{p}}<{\mathrm{K}}_{\mathrm{p}}$the reaction proceeds towards the right side, favouring the reactants until the equilibrium is attained.

When,${\mathrm{Q}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{p}}$ no net charge takes place since the reaction is at equilibrium.

We have,

$$\begin{gathered} {\mathrm{P}}_{{\mathrm{H}}_2}=0.010\mathrm{atm} \\ {\mathrm{P}}_{{\mathrm{Br}}_2}=0.010\mathrm{atm} \\ {\mathrm{P}}_{\mathrm{HBr}}=0.20\mathrm{atm} \end{gathered}$$

Substitute the initial partial pressure of all the species in the 1^st equation to calculate the value oflocalid="1655097112320" ${\mathrm{Q}}_p$

The reaction quotient of the reaction is,

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{p}}=\frac{\left({\mathrm{P}}_{{\mathrm{H}}_2}\right)\left({\mathrm{P}}_{{\mathrm{Br}}_2}\right)}{{\left({\mathrm{P}}_{\mathrm{HBr}}\right)}^2} \\ =\frac{(0.010)(0.010)}{{(0.20)}^2} \\ =2.5\times {10}^{-3} \end{gathered}$$

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{P}}=\frac{\left({\mathrm{P}}_{{\mathrm{H}}_2}\right)\left({\mathrm{P}}_{{\mathrm{Br}}_2}\right)}{{\left({\mathrm{P}}_{\mathrm{HBr}}\right)}^2} \\ =\frac{(0.010)(0.010)}{(0.20)} \\ =2.5\times {10}^{-3} \end{gathered}$$

Therefore, the reaction quotient ${\mathrm{Q}}_{\mathrm{p}}$of the reaction is $2.5\times {10}^{-3}.$

The reaction quotient ${\mathrm{Q}}_{\mathrm{p}}$of the reaction is$2.5\times {10}^{-3}$

The equilibrium constant ${\mathrm{K}}_{\mathrm{p}}$of the reaction is $4.18\times {10}^{-9}$$4.18\times {10}^{-9}$.

Since the reaction proceeds towards the left side, favouring the reactants until the equilibrium is attained.

Therefore, the reaction is not at equilibrium and the reaction proceeds towards the left side, favouring the reactants until the equilibrium is attained.