MHT CET · Chemistry · Chemical Kinetics
The rate law for the reaction \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}\) at \(25^{\circ} \mathrm{C}\) is given by rate \(=k[A][B]^2\). Calculate the rate of reaction if rate constant at same temperature is \(6.25 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1}[[\mathrm{~A}]=1 \mathrm{M},[\mathrm{B}]=0.2 \mathrm{M}]\)
- A \(0.25 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
- B \(0.5 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
- C \(0.75 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
- D \(1.25 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(A) \(0.25 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \text { Rate } & =\mathrm{k}[\mathrm{A}][\mathrm{B}]^2 \\ & =6.25 \mathrm{~mol}^{-2} \mathrm{dm}^6 \mathrm{~s}^{-1} \times 1 \mathrm{~mol} \mathrm{dm}^{-3} \\ & \quad \times 0.2 \mathrm{~mol} \mathrm{dm}^{-3} \times 0.2 \mathrm{~mol} \mathrm{dm}^{-3} \\ & =0.25 \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}\end{aligned}\)
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