JEE MAIN - Chemistry (2024 - 30th January Evening Shift - No. 27)
$$\mathrm{NO}_2$$ required for a reaction is produced by decomposition of $$\mathrm{N}_2 \mathrm{O}_5$$ in $$\mathrm{CCl}_4$$ as by equation
$$2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$$
The initial concentration of $$\mathrm{N}_2 \mathrm{O}_5$$ is $$3 \mathrm{~mol} \mathrm{~L}^{-1}$$ and it is $$2.75 \mathrm{~mol} \mathrm{~L}^{-1}$$ after 30 minutes.
The rate of formation of $$\mathrm{NO}_2$$ is $$\mathrm{x} \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$$, value of $$\mathrm{x}$$ is _________. (nearest integer)
Explanation
Rate of reaction (ROR)
$$\begin{aligned} & =-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=\frac{1}{4} \frac{\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=\frac{\Delta\left[\mathrm{O}_2\right]}{\Delta \mathrm{t}} \\ & \mathrm{ROR}=-\frac{1}{2} \frac{\Delta\left[\mathrm{N}_2 \mathrm{O}_5\right]}{\Delta \mathrm{t}}=-\frac{1}{2} \frac{(2.75-3)}{30} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \\ & \mathrm{ROR}=-\frac{1}{2} \frac{(-0.25)}{30} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \\ & \text { ROR }=\frac{1}{240} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \end{aligned}$$
Rate of formation of $$\mathrm{NO}_2=\frac{\Delta\left[\mathrm{NO}_2\right]}{\Delta \mathrm{t}}=4 \times \mathrm{ROR}$$
$$=\frac{4}{240}=16.66 \times 10^{-3} \mathrm{molL}^{-1} \mathrm{~min}^{-1} \simeq 17 \times 10^{-3} \text {. }$$
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