MHT CET · Chemistry · Chemical Kinetics

For the reaction
\(2 \mathrm{NOBr}_{(\mathrm{g})} \rightarrow 2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{Br}_{ \left.2_{(\mathrm{g})}\right)}\), rate law is \(\mathrm{r}=\mathrm{K}[\mathrm{NOBr}]^{2}\)
If rate constant is \(1 \cdot 62 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and concentration of \(\mathrm{NOBr}\) is \(2 \cdot 00 \times 10^{-3} \mathrm{M}\), What is the rate of reaction?

A \(6 \cdot 48 \times 10^{-6} \mathrm{Ms}^{-1}\)
B \(4 \cdot 05 \times 10^{-5} \mathrm{Ms}^{-1}\)
C \(2 \cdot 46 \times 10^{-6} \mathrm{Ms}^{-1}\)
D \(5 \cdot 24 \times 10^{-6} \mathrm{Ms}^{-1}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(6 \cdot 48 \times 10^{-6} \mathrm{Ms}^{-1}\)

Step-by-step Solution

Detailed explanation

\(2 \mathrm{NOBr}_{(\mathrm{g})} \rightarrow 2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{Br}_{2(\mathrm{~g})}\),
\(\mathrm{k}=1.62 \mathrm{~M}^{-1} \mathrm{~s}^{-1},[\mathrm{NOBr}]=2.00 \times 10^{-3} \mathrm{~M}\)
Now, \(\mathrm{r}=\mathrm{k}[\mathrm{NOBr}]^{2}\)
\(\therefore \mathrm{r}=1.62 \mathrm{~M}^{-1} \mathrm{~s}^{-1} \times\left(2.00 \times 10^{-3}\right)^{2} \mathrm{M}^{2}\)
\(\therefore \mathrm{r}=6.48 \times 10^{-6} \mathrm{~M} \mathrm{~s}^{-1}\)

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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