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For the reaction \mathrm{ \mathrm{N}_2 \mathrm{O}_4 \rightarrow 2 \mathrm{NO}_2,} rate constant is \mathrm{2.48 \times 10^{-4} \mathrm{~s}^{-1}} Find molar ratio of\mathrm{2.48 \times 10^{-4} \mathrm{~s}^{-1}}\mathrm{ \mathrm{NO}_2 to \mathrm{N}_2 \mathrm{O}_4~ after ~27.25 \mathrm{~min}= }?

Option: 1

1


Option: 2

2


Option: 3

3


Option: 4

4


Answers (1)

best_answer

\begin{aligned} &\mathrm{N}_2 \mathrm{O}_4 \rightarrow 2 \mathrm{NO}_2\\ &\begin{array}{cccc} \text { Initial } & \mathrm{a} & - \\ \mathrm{t} & \mathrm{a-x} & - \end{array}\\ \mathrm{k} & =\mathrm{\frac{2.303}{t} \log \frac{a}{a-x} }\\ \mathrm{2.48 \times 10^{-4}} & =\mathrm{\frac{2.303}{27.25 \times 60} \log \frac{a}{a-x}} \\ &\mathrm{ \frac{a}{a-x}=10^{0.176}=1.5} \\ &\mathrm{ x=\frac{a }{3}} \\ &\mathrm{ \frac{n_{\mathrm{NO}_2}}{n_{\mathrm{N}_2 \mathrm{O}_4}}}=\mathrm{\frac{2 x}{a-x}=\frac{2 a / 3}{2 a / 3}=1 }\end{aligned}

 

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Divya Prakash Singh

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