Q55 / 180 · COMEDK 2024 (Memory Based) · Chemistry
The equilibrium constant of given reaction is 100 .
$\mathrm{N}_2(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2$ (gas)
What is the equilibrium constant for the reaction given below?
$\mathrm{NO}_2(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(g)+\mathrm{O}_2(g)$
$\mathrm{N}_2(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2$ (gas)
What is the equilibrium constant for the reaction given below?
$\mathrm{NO}_2(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(g)+\mathrm{O}_2(g)$
- A10
- B1
- Correct answerC0.1
- D0.01
Solution
$\mathrm{N}_2+2 \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}_2 ; k=100$...(i)
To find equilibrium constant of
$\mathrm{NO}_2(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(g)+\mathrm{O}_2(g)$...(ii)
If we divide eq (i) by 2 and taking the reverse we will obtain equation (ii).
$\begin{aligned}<br/>\text { Required } k & =\sqrt{\frac{1}{100}}=\frac{1}{10}=0.1 \\<br/>k & =0.1<br/>\end{aligned}$
To find equilibrium constant of
$\mathrm{NO}_2(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(g)+\mathrm{O}_2(g)$...(ii)
If we divide eq (i) by 2 and taking the reverse we will obtain equation (ii).
$\begin{aligned}<br/>\text { Required } k & =\sqrt{\frac{1}{100}}=\frac{1}{10}=0.1 \\<br/>k & =0.1<br/>\end{aligned}$
Hence, the correct answer is option: 'C'.