The rate for the reaction, $RCl + NaOH\left( {aq} \right) \to ROH + NaCl$ is given by $rate{\rm{ }} = K_1\left[ {RCl} \right].$ The rate of the reaction is

Unaffected by increasing the temperature of the reaction

Decreased on increasing the temperature of the reaction

Doubled on doubling the concentration of $NaOH$

JEE CHEMISTRY MCQ: CHEMICAL KINETICS

1 Medium

1 of 1

Understanding +4 | -1

The rate for the reaction,

RCl + NaOH\left( {aq} \right) \to ROH + NaCl

is given by

rate\mathrm{ } = K_1\left[ {RCl} \right].

The rate of the reaction is

View answer and explanation

Correct option: C

Halved on reducing the concentration of

RCl

to half

The correct answer is C (Halved on reducing the concentration of

RCl

to half) because the rate law is first order with respect to

RCl

, so halving its concentration will halve the rate.

Practice All Chapter MCQs →

\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