JEE CHEMISTRY MCQ: CHEMICAL KINETICS

1 Hard

1 of 1

Analyzing +4 | -1

Consider the two hypothetical reactions given below:

I

a A \rightarrow

Products,

k=x \mathrm{~mol}^{-1} \mathrm{~L} \mathrm{~min}^{-1}

b B \rightarrow

Products,

k=y \mathrm{~min}^{-1}

The half-lives of both the reactions are the same, equal to

1 \mathrm{hr}

when molar concentration of the reactant is

1.0 \mathrm{M}

in each case. If these reactions are started at the same time taking

1 \mathrm{M}

of the reactant in each case, the ratio

[A]/[B]

after

3 \mathrm{hr}

will be:

View answer and explanation

Correct option: D

Option D is correct. The ratio [A]/[B] after 3 hours is 2, calculated by determining the order of each reaction and then finding the concentration of A and B after 3 hours.

More Questions from CHEMICAL KINETICS

Practice All Chapter MCQs →

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

)