(B) The given rate law is $\text{Rate} = K_1[RCl]$.This indicates that the reaction is of first order with respect to the alkyl halide $(RCl)$ and zer

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Halved on reducing the concentration of alkyl halide to one half

(B) The given rate law is $\text{Rate} = K_1[RCl]$.This indicates that the reaction is of first order with respect to the alkyl halide $(RCl)$ and zero order with respect to sodium hydroxide $(NaOH)$.If the concentration of the alkyl halide is reduced to one half,the new rate will be $\text{Rate}' = K_1 \times \frac{1}{2}[RCl] = \frac{1}{2} \times \text{Rate}$.Therefore,the rate is halved on reducing the concentration of the alkyl halide to one half.

Class 12 Chemistry Chemical Kinetics Rate law , Rate constant , Order of Reaction and Molecularity

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The rate law for the reaction $RCl + NaOH_{(aq)} \to ROH + NaCl$ is given by $\text{Rate} = K_1[RCl]$. The rate of the reaction will be

IIT 1988 All IIT

A
Doubled on doubling the concentration of sodium hydroxide
B
Halved on reducing the concentration of alkyl halide to one half
C
Decreased on increasing the temperature of the reaction
D
Unaffected by increasing the temperature of the reaction

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

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The reaction,$X + 2Y + Z \to N$ occurs by the following mechanism:
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The rate law for the reaction $RCl + NaOH_{(aq)} \to ROH + NaCl$ is given by $\text{Rate} = K_1[RCl]$. The rate of the reaction will be

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The reaction,$X + 2Y + Z \to N$ occurs by the following mechanism:
$(i)$ $X + Y \rightleftharpoons M$ (very rapid equilibrium)
$(ii)$ $M + Z \to P$ (slow)
$(iii)$ $P + Y \to N$ (very fast)
What is the rate law for this reaction?