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JEE CHEMISTRY MCQ: CHEMICAL KINETICS

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Analyzing +4 | -1

Assertion (A):A plot of

\log \frac{d X}{d t} \text{ vs. }\log (a-x)

leads to value of

K

as antilog (intercept) and slope equal to order of reaction.

Reason (R):

\frac{d X}{d t}=K(a-x)^n

View answer and explanation

Correct option: C

Both A and R are correct and R is correct explanation of A

Option D (Both A and R are correct but R is not correct explanation of A) is incorrect. Option C (Both A and R are correct and R is correct explanation of A) is correct because the integrated rate law can be expressed in logarithmic form, where the intercept corresponds to the rate constant and the slope corresponds to the order of the reaction.

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\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 \%

\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