Q35P - Chemistry: Molecular Nature Of Matter And Change

Question

 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

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

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?

Step-by-Step Solution

Answer

Since ${\mathrm{Q}}_{\mathrm{p}}>{\mathrm{K}}_{\mathrm{p}}$, 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.

Step 1: Write the equilibrium equation

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.

Step 2: Is the reaction 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 of ${\mathrm{Q}}_p$

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.