JEE CHEMISTRY MCQ: CHEMICAL KINETICS

1 Hard

1 of 1

Application +4 | -1

For a reaction of order

n,

the integrated form of the rate equation is

(n-1) \cdot K \cdot t=\left(C_0\right)^{1-n}-(C)^{1-n},

where

C_0

and

C

are the values of the reactant concentration at the start and after time '

t

'. What is the relationship between

t_{3 / 4}

and

t_{1 / 2},

where

t_{3 / 4}

is the time required for

C

to become

C_0 / 4.

View answer and explanation

Correct option: A

t_{3 / 4}=t_{1 / 2} \cdot\left[2^{n-1}+1\right]

The correct answer is

t_{3 / 4}=t_{1 / 2} \cdot\left[2^{n-1}+1\right]

because it is derived from the integrated rate equation for a reaction of order n.

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