JEE CHEMISTRY MCQ: CHEMICAL KINETICS

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Hydrolysis of ethyl acetate is catalysed by acid.

\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5+\mathrm{H}_2 \mathrm{O} \xrightarrow{\mathrm{H}^{+}} \mathrm{CH}_3 \mathrm{COOH}+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}

Rate constant of the reaction is

5.56 \times 10^{-6} \mathrm{~s}^{-1}

. What is its rate constant when it is second-order reaction (pseudo unimolecular)?

View answer and explanation

Correct option: C

1.0 \times 10^{-7} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}

The correct answer is

1.0 \times 10^{-7} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}

because it is calculated by dividing the pseudo-first-order rate constant by the concentration of water.

More Questions from CHEMICAL KINETICS

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\mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB}

\begin{aligned} & {{\text{A}}_2}\mathop \rightleftharpoons \limits_{{{\text{k}}_{ - 1}}}^{{{\text{k}}_1}} {\text{A}} + {\text{A (fast) }} \\ & \mathrm{A}+\mathrm{B}_2 \xrightarrow{\mathrm{k}_2} \mathrm{AB}+\mathrm{B} \text { (slow) } \\ & \mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB} \text { (fast) } \end{aligned}

k

(Initial\mathrm{ }conc.\mathrm{ } = a)

C_0 \sqrt{T}

C_0

T

100 \mathrm{~kJ/mol}.

75 \%

\log _{10} \frac{K_2}{K_1}

27^{\circ} \mathrm{C}

2.303 \times 8.314=19.147