MHT CET · Chemistry · Chemical Kinetics
In the reaction, \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3}\), the rate of disappearance of \(\mathrm{H}_{2}\) is \(0 \cdot 02 \mathrm{M} / \mathrm{s}\). The rate of appearance of \(\mathrm{NH}_{3}\) is
- A \(0 \cdot 0133 \mathrm{M} / \mathrm{s}\)
- B \(0 \cdot 023 \mathrm{M} / \mathrm{s}\)
- C \(0 \cdot 004 \mathrm{M} / \mathrm{s}\)
- D \(0 \cdot 032 \mathrm{M} / \mathrm{s}\)
Answer & Solution
Correct Answer
(A) \(0 \cdot 0133 \mathrm{M} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3}\)
Rate of reaction \(=-\frac{\mathrm{d}\left(\mathrm{N}_{2}\right)}{\mathrm{dt}}=-\frac{1}{3} \frac{\mathrm{d}\left(\mathrm{H}_{2}\right)}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{d}\left(\mathrm{NH}_{3}\right)}{\mathrm{dt}}\)
From rate expression,
\(\frac{\mathrm{d}\left(\mathrm{NH}_{2}\right)}{\mathrm{dt}}=\frac{2}{3} \frac{\mathrm{d}\left(\mathrm{H}_{2}\right)}{\mathrm{d} \mathrm{t}}=\frac{2}{3} \times 0.02=0.0133 \mathrm{M} / \mathrm{s}\)
Rate of reaction \(=-\frac{\mathrm{d}\left(\mathrm{N}_{2}\right)}{\mathrm{dt}}=-\frac{1}{3} \frac{\mathrm{d}\left(\mathrm{H}_{2}\right)}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{d}\left(\mathrm{NH}_{3}\right)}{\mathrm{dt}}\)
From rate expression,
\(\frac{\mathrm{d}\left(\mathrm{NH}_{2}\right)}{\mathrm{dt}}=\frac{2}{3} \frac{\mathrm{d}\left(\mathrm{H}_{2}\right)}{\mathrm{d} \mathrm{t}}=\frac{2}{3} \times 0.02=0.0133 \mathrm{M} / \mathrm{s}\)
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