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Equilibrium: JEE Main Chemistry Question with Solution

For reaction:

\mathrm{SO}_2(\mathrm{g}) + \frac{1}{2}\mathrm{O}_2(\mathrm{g}) \rightleftharpoons \mathrm{SO}_3(\mathrm{g})

\mathrm{K}_\mathrm{p} = 2 \times 10^{12}

at

27^\circ\mathrm{C}

and

1\text{ atm}

pressure. The

\mathrm{K}_\mathrm{c}

for the same reaction is

\times 10^{13}

. (Nearest integer)

(\text{Given } \mathrm{R} = 0.082\text{ L atm K}^{-1}\text{ mol}^{-1}\text{ ) }

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Step-by-step solutionView

Correct answer

The value of $K_c$ is $1 imes 10^{13}$ , so the integer to fill in is 1.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

For the reaction $\mathrm{SO}_2(\mathrm{g}) + \frac{1}{2}\mathrm{O}_2(\mathrm{g}) \rightleftharpoons \mathrm{SO}_3(\mathrm{g})$ : $K_p = 2 \times 10^{12}$ , $T = 27^\circ\mathrm{C} = 300\text{ K}$ , $R = 0.082\text{ L atm K}^{-1}\text{ mol}^{-1}$ .

02find

The pre-factor $x$ in $K_c = x \times 10^{13}$ to the nearest integer.

03strategise

Calculate the change in gaseous moles: $\Delta n_g = n_{\text{products}} - n_{\text{reactants}} = 1 - \left(1 + \frac{1}{2}\right) = -\frac{1}{2}$ . Then apply the relation $K_p = K_c(RT)^{\Delta n_g}$ , which rearranges to $K_c = K_p(RT)^{-\Delta n_g} = K_p(RT)^{1/2}$ .

04execute

First, evaluate the product $RT$ : $RT = 0.082 \times 300 = 24.6$ . Then calculate $(RT)^{1/2} = \sqrt{24.6} \approx 4.96$ . Now evaluate $K_c$ : $K_c = K_p \times \sqrt{RT} = (2 \times 10^{12}) \times 4.96 = 9.92 \times 10^{12} = 0.992 \times 10^{13} \approx 1 \times 10^{13}$ . Thus, the integer pre-factor is 1.

✓verify

Since $\sqrt{24.6} \approx 5$ , $K_c \approx 2 \times 10^{12} \times 5 = 10 \times 10^{12} = 1 \times 10^{13}$ . The nearest integer is unambiguously 1.

Hints that build this answer step by step

What is the change in the number of moles of gas, $\Delta n_g$ , for the given reaction?
$\Delta n_g = 1 - \left(1 + \frac{1}{2}\right) = -0.5$
Which equation correctly relates $K_c$ to $K_p$ ?
$K_c = K_p (RT)^{-\Delta n_g}$
Evaluating $K_c = 2 \times 10^{12} \times (0.082 \times 300)^{0.5}$ , what is the numerical multiplier for $10^{13}$ rounded to the nearest integer?
1

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Chemistry
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Does total pressure (1 atm) affect the value of Kc?

No, equilibrium constants Kp and Kc depend only on temperature for ideal gas mixtures; the pressure of 1 atm is just additional contextual information.

Answer

The value of $K_c$ is $1 imes 10^{13}$ , so the integer to fill in is 1.

Why each option works or fails

Step-by-step solution

given: For the reaction $\mathrm{SO}_2(\mathrm{g}) + \frac{1}{2}\mathrm{O}_2(\mathrm{g}) \rightleftharpoons \mathrm{SO}_3(\mathrm{g})$ : $K_p = 2 \times 10^{12}$ , $T = 27^\circ\mathrm{C} = 300\text{ K}$ , $R = 0.082\text{ L atm K}^{-1}\text{ mol}^{-1}$ .
find: The pre-factor $x$ in $K_c = x \times 10^{13}$ to the nearest integer.
strategise: Calculate the change in gaseous moles: $\Delta n_g = n_{\text{products}} - n_{\text{reactants}} = 1 - \left(1 + \frac{1}{2}\right) = -\frac{1}{2}$ . Then apply the relation $K_p = K_c(RT)^{\Delta n_g}$ , which rearranges to $K_c = K_p(RT)^{-\Delta n_g} = K_p(RT)^{1/2}$ .
execute: First, evaluate the product $RT$ : $RT = 0.082 \times 300 = 24.6$ . Then calculate $(RT)^{1/2} = \sqrt{24.6} \approx 4.96$ . Now evaluate $K_c$ : $K_c = K_p \times \sqrt{RT} = (2 \times 10^{12}) \times 4.96 = 9.92 \times 10^{12} = 0.992 \times 10^{13} \approx 1 \times 10^{13}$ . Thus, the integer pre-factor is 1.
verify: Since $\sqrt{24.6} \approx 5$ , $K_c \approx 2 \times 10^{12} \times 5 = 10 \times 10^{12} = 1 \times 10^{13}$ . The nearest integer is unambiguously 1.