JEE CHEMISTRY MCQ: CHEMICAL KINETICS

1 Hard

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Analyzing +4 | -1

In the reaction

a \mathrm{~A}+b \mathrm{~B}+c \mathrm{C} \rightarrow

Products,

(I) if concentration of

\mathrm{A}

is doubled, keeping concentration of

\mathrm{B}

and

\mathrm{C}

constant, the rate of reaction becomes doubled.
(II) if concentration of

B

is halved, concentration of

A

and

C

constant, the rate of reaction remains unaffected.

(III) if concentration of

\mathrm{C}

is made 1.5 times, the rate of reaction becomes 2.25 times.

The order of reaction is

View answer and explanation

Correct option: C

Option C (3) is correct. The rate law is determined to be Rate = k[A][C]², leading to an overall order of 3.

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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