JEE CHEMISTRY MCQ: CHEMICAL KINETICS

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Analyzing +4 | -1

Consider the following statements,

The rate law for the acid catalysed hydrolysis of an ester being given as

Rate = k\;\left[ {{H^ + }} \right]\;\left[ {ester} \right] = k'\;\left[ {ester} \right]

If the acid concentration is doubled at constant ester concentration

1. The second order rate constant,

k

is doubled.

2. The pseudo first order rate constant,

k

is double.

3. The rate of the reaction is doubled.

Which of the above statements are correct?

View answer and explanation

Correct option: B

2 and 3

Option B (2 and 3) is correct because doubling the acid concentration doubles both the pseudo-first-order rate constant and the reaction rate, while the second-order rate constant remains unchanged.

More Questions from CHEMICAL KINETICS

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The reaction

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

follows the mechanism

\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}

The overall order of the reaction is: For the first order reaction with the rate constant

k

, which expression gives the rate half-life period?

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

Which of the following is a zero order reaction? For a certain reaction involving a single reactant, it is found that

C_0 \sqrt{T}

is constant, where

C_0

is the initial concentration of the reactant and

T

is the half-life. What is the order of the reaction? Arrhenius equation is The energy of activation for an uncatalysed reaction is

100 \mathrm{~kJ/mol}.

Presence of a catalyst lowers the energy of activation by

75 \%

. The

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

of the ratio of rate constant of catalysed and uncatalysed reactions at

27^{\circ} \mathrm{C}

is?

Assume the frequency factor is same for both reactions. (Given

2.303 \times 8.314=19.147

)