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$9.2 \mathrm{~g} \mathrm{~N}_2 \mathrm{O}_4$ is heated in a 1 L vessel till equilibrium state is established
$\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$
In equilibrium state $50 \% \mathrm{~N}_2 \mathrm{O}_4$ was dissociated, equilibrium constant will be (mol. wt. of $\mathrm{N}_2 \mathrm{O}_4=92$ )
Options:
$\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$
In equilibrium state $50 \% \mathrm{~N}_2 \mathrm{O}_4$ was dissociated, equilibrium constant will be (mol. wt. of $\mathrm{N}_2 \mathrm{O}_4=92$ )
Solution:
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Verified Answer
The correct answer is:
0.2
$\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})$
Molar concentration of
$\left[\mathrm{N}_2 \mathrm{O}_4\right]=\frac{9.2}{92}=0.1 \mathrm{~mol} / \mathrm{L}$
In equilibrium state when it 50\% dissociates,
$\begin{aligned}
{\left[\mathrm{N}_2 \mathrm{O}_4\right] } & =0.05 \mathrm{M} \\
{\left[\mathrm{NO}_2\right] } & =0.1 \mathrm{M} \\
K_c & =\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2 \mathrm{O}_4\right]} \\
& =\frac{0.1 \times 0.1}{0.05}=0.2
\end{aligned}$
Molar concentration of
$\left[\mathrm{N}_2 \mathrm{O}_4\right]=\frac{9.2}{92}=0.1 \mathrm{~mol} / \mathrm{L}$
In equilibrium state when it 50\% dissociates,
$\begin{aligned}
{\left[\mathrm{N}_2 \mathrm{O}_4\right] } & =0.05 \mathrm{M} \\
{\left[\mathrm{NO}_2\right] } & =0.1 \mathrm{M} \\
K_c & =\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2 \mathrm{O}_4\right]} \\
& =\frac{0.1 \times 0.1}{0.05}=0.2
\end{aligned}$
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