MHT CET · Chemistry · Chemical Kinetics

Ammonia and oxygen react at high temperature as in reaction, \(4 \mathrm{NH}_{3(g)}+5 \mathrm{O}_{2(g)} \rightarrow 4 \mathrm{NO}_{(g)}+6 \mathrm{H}_2 \mathrm{O}_{(g)}\) If rate of formation of \(\mathrm{NO}\) is \(3.6 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{sec}^{-1}\). Calculate the rate of formation of water.

A \(5.4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{sec}^{-1}\)
B \(6.0 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-3} \mathrm{sec}^{-1}\)
C \(1.8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{sec}^{-1}\)
D \(3.6 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{sec}^{-1}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(5.4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{sec}^{-1}\)

Step-by-step Solution

Detailed explanation

\(-\frac{1}{4} \frac{\mathrm{d}\left[\mathrm{NH}_3\right]}{\mathrm{d} t}=+\frac{1}{4} \frac{\mathrm{d}[\mathrm{NO}]}{\mathrm{d} t}=\frac{1}{6} \frac{\mathrm{d}\left[\mathrm{H}_2 \mathrm{O}\right]}{\mathrm{d} t}\)
\(\begin{aligned} & \frac{\mathrm{d}\left[\mathrm{H}_2 \mathrm{O}\right]}{\mathrm{d} t}=\frac{6}{4} \frac{\mathrm{d}[\mathrm{NO}]}{\mathrm{d} t}=\frac{3}{2} \times 3.6 \times 10^{-3} \\ & =5.4 \times 10^{-3} \frac{\mathrm{mol}}{\mathrm{L}-\mathrm{sec}}\end{aligned}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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