Question

The value of K_p for the following reaction $2{\mathrm{SO}}_2\left(\mathrm{g}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{SO}}_3\left(\mathrm{g}\right)$is $5$. What will be the partial pressure of ${\mathrm{O}}_2$ at equilibrium when equal moles of ${\mathrm{SO}}_3$and ${\mathrm{SO}}_2$ are present at equilibrium?

• A. $0.1$
• B. $0.2$
• C. $0.3$
• D. $0.5$

Answer 1

The correct option is B

$0.2$

Explanation for correct option

(B)$0.2$

Step1. Expression for K_p

$K_p=\frac{{\left[p_{{\text{SO}}_3}\right]}^2}{{\left[{\mathit{\text{p}}}_{{\text{SO}}_{\text{2}}}\right]}^2\left[{\mathit{\text{p}}}_{{\text{O}}_2}\right]}$

Step2. Calculate the partial pressure of ${\mathrm{O}}_2$

$5=\frac{{\left[p_{{\text{SO}}_3}\right]}^2}{{\left[{\mathit{\text{p}}}_{{\text{SO}}_{\text{2}}}\right]}^2\left[{\mathit{\text{p}}}_{{\text{O}}_2}\right]}$

As $\left[{\mathit{\text{p}}}_{{\text{SO}}_{\text{2}}}\right]=\left[p_{{\text{SO}}_3}\right]$, So, the above equation is simplified as follows:-

$$\begin{gathered} 5=\frac{1}{{\mathit{\text{p}}}_{{\text{O}}_2}} \\ {\mathit{\text{p}}}_{{\text{O}}_2}\mathit{\text{=}}\frac{1}{5} \\ \mathit{\text{=}}0.2 \end{gathered}$$

Hence, the partial pressure of ${\mathrm{O}}_2$ at equilibrium when equal moles of ${\mathrm{SO}}_3$and ${\mathrm{SO}}_2$ are present at equilibrium is $0.2$.

Explanation for incorrect option

The value of the partial pressure of ${\mathrm{O}}_2$ at equilibrium is $0.2$. Hence, options (A), (C), and (D) are incorrect.

Hence, option (B) is correct. The value of the partial pressure of ${\mathrm{O}}_2$ at equilibrium is $0.2$.