MHT CET · Chemistry · Chemical Kinetics
The reaction \(2 \mathrm{~A}+\mathrm{B}+\mathrm{C} \longrightarrow \mathrm{D}+\mathrm{E}\) is found to. be first order in A , second order in B and zero order in C . What is the effect of increasing concentration of all reactants twice?
- A Rate of reaction increases 8 times.
- B Rate of reaction increases 24 times.
- C Rate of reaction increases 36 times.
- D Rate of reaction remains unaffected.
Answer & Solution
Correct Answer
(A) Rate of reaction increases 8 times.
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { Rate }=\mathrm{k}[\mathrm{~A}][\mathrm{B}]^2[\mathrm{C}]^0 \\
& (\text { Rate })_1=\mathrm{k} \times[2 \mathrm{~A}] \times[2 \mathrm{~B}]^2 \times[2 \mathrm{C}]^0 \\
& =8 \mathrm{k}[\mathrm{~A}][\mathrm{B}]^2[\mathrm{C}]^0 \\
& \therefore \quad \frac{(\text { Rate })_1}{\text { Rate }}=\frac{8 \mathrm{k}[\mathrm{~A}][\mathrm{B}]^2[\mathrm{C}]^0}{\mathrm{k}[\mathrm{~A}][\mathrm{B}]^2[\mathrm{C}]^0}=8 \\
& \therefore \quad(\text { Rate })_1=8 \times \text { Rate }
\end{aligned}\)
i.e., rate of reaction increases 8 times.
& \text { Rate }=\mathrm{k}[\mathrm{~A}][\mathrm{B}]^2[\mathrm{C}]^0 \\
& (\text { Rate })_1=\mathrm{k} \times[2 \mathrm{~A}] \times[2 \mathrm{~B}]^2 \times[2 \mathrm{C}]^0 \\
& =8 \mathrm{k}[\mathrm{~A}][\mathrm{B}]^2[\mathrm{C}]^0 \\
& \therefore \quad \frac{(\text { Rate })_1}{\text { Rate }}=\frac{8 \mathrm{k}[\mathrm{~A}][\mathrm{B}]^2[\mathrm{C}]^0}{\mathrm{k}[\mathrm{~A}][\mathrm{B}]^2[\mathrm{C}]^0}=8 \\
& \therefore \quad(\text { Rate })_1=8 \times \text { Rate }
\end{aligned}\)
i.e., rate of reaction increases 8 times.
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