MHT CET 2025 20th April Morning Shift | Chemical Kinetics Question 1 | Chemistry

The rate constant for the reaction, $2 \mathrm{~N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{~N}_2 \mathrm{O}_{4(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}$ is $4.98 \times 10^{-4} \mathrm{~s}^{-1}$. What is the order of reaction?

Zero

MHT CET 2025 20th April Morning Shift

MCQ (Single Correct Answer)

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Which of the following equations is correct regarding rate of disappearance of reactant and appearance of product for

$$ \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \longrightarrow 2 \mathrm{NH}_{3(\mathrm{~g})} $$

$3 \frac{\mathrm{~d}\left[\mathrm{~N}_2\right]}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{~N}_2\right]}{\mathrm{dt}}$

$\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{~N}_2\right]}{\mathrm{dt}}=\frac{1}{3} \frac{\mathrm{~d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}$

$2 \frac{\mathrm{~d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}}=3 \frac{\mathrm{~d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}$

$\quad 3 \frac{\mathrm{~d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}}=-2 \frac{\mathrm{~d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}$

MHT CET 2025 19th April Evening Shift

MCQ (Single Correct Answer)

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Which from following is a correct representation of reaction rate for reaction stated below?

$$ \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})} $$

$\frac{\mathrm{d}\left[\mathrm{N}_2\right]}{\mathrm{dt}}=-3 \frac{\mathrm{~d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=2 \frac{\mathrm{~d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}}$

$\frac{\mathrm{d}\left[\mathrm{N}_2\right]}{\mathrm{dt}}=-\frac{1}{3} \frac{\mathrm{~d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}}$

$-\frac{\mathrm{d}\left[\mathrm{N}_2\right]}{\mathrm{dt}}=-\frac{1}{3} \frac{\mathrm{~d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=\frac{\mathrm{1}}{2} \frac{\mathrm{~d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}}$

$\frac{\mathrm{d}\left[\mathrm{N}_2\right]}{\mathrm{dt}}=\frac{1}{3} \frac{\mathrm{~d}\left[\mathrm{H}_2\right]}{\mathrm{dt}}=-\frac{1}{2} \frac{\mathrm{~d}\left[\mathrm{NH}_3\right]}{\mathrm{dt}}$

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