(B) Let the rate law be $\text{Rate} = K[A]^x[B]^y$.Using data $1$ and $4$ (where $[B]$ is constant):$\frac{3.0 \times 10^{-2}}{7.5 \times 10^{-3}} =

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(B) Let the rate law be $\text{Rate} = K[A]^x[B]^y$.Using data $1$ and $4$ (where $[B]$ is constant):$\frac{3.0 \times 10^{-2}}{7.5 \times 10^{-3}} = (\frac{0.4}{0.1})^x$ $\Rightarrow 4 = 4^x$ $\Rightarrow x = 1$.Using data $2$ and $3$ (where $[A]$ is constant):$\frac{3.6 \times 10^{-1}}{9.0 \times 10^{-2}} = (\frac{0.4}{0.2})^y$ $\Rightarrow 4 = 2^y$ $\Rightarrow y = 2$.Thus,the rate law is $\text{Rate} = K[A][B]^2$.

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

Easy

For the reaction $2A + B \rightarrow C + D$,select the correct rate law based on the following data:
$1$. $[A] = 0.1, [B] = 0.1, \text{Initial Rate} = 7.5 \times 10^{-3}$
$2$. $[A] = 0.3, [B] = 0.2, \text{Initial Rate} = 9.0 \times 10^{-2}$
$3$. $[A] = 0.3, [B] = 0.4, \text{Initial Rate} = 3.6 \times 10^{-1}$
$4$. $[A] = 0.4, [B] = 0.1, \text{Initial Rate} = 3.0 \times 10^{-2}$

A
Rate $= K[A]^2[B]$
B
Rate $= K[A][B]^2$
C
Rate $= K[A][B]^3$
D
Rate $= K[A][B]$

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Chemical Kinetics

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Rate law , Rate constant , Order of Reaction and Molecularity

For the reaction $Cl_{2(aq)} + H_2S_{(aq)} \to S_{(s)} + 2H^+_{(aq)} + 2Cl^-_{(aq)}$, the rate law is given by $\text{Rate} = K[Cl_2][H_2S]$. Which of the following mechanisms is consistent with this rate law?
$(A)$ $Cl_2 + H_2S \to H^+ + Cl^- + Cl^+ + HS^-$ (slow); $Cl^+ + HS^- \to H^+ + Cl^- + S$ (fast)
$(B)$ $H_2S \rightleftharpoons H^+ + HS^-$ (fast equilibrium); $Cl_2 + HS^- \to 2Cl^- + H^+ + S$ (slow)

Medium

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The conversion of molecules $X$ to $Y$ follows second order kinetics. If the concentration of $X$ is increased to three times,how will it affect the rate of formation of $Y$?

Medium

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The mechanism of reaction $2 \ NO + Cl_2 \rightarrow 2 \ NOCl$ is given as :
$(i) 2 NO \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} N_2 O _2$ (fast)
$(ii) N_2O_2 + Cl_2 \xrightarrow{K_3} 2 \ NOCl$ (slow)
The rate expression of the reaction will be :

Difficult

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When the concentration of a reactant,$A$,in a reaction $A \to \text{Products}$ is doubled,the rate of reaction increases seven times. The order of the reaction is between:

Medium

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For the reaction $A + 2B \rightarrow \text{Product } (P)$,the rate law is given by $\frac{d[P]}{dt} = K[A]^2[B]$. If $[A]$ is taken in large excess,what will be the order of the reaction?

Medium

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The conversion of molecules $X$ to $Y$ follows second order kinetics. If the concentration of $X$ is increased to three times,how will it affect the rate of formation of $Y$?

The mechanism of reaction $2 \ NO + Cl_2 \rightarrow 2 \ NOCl$ is given as : $(i) 2 NO \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} N_2 O _2$ (fast) $(ii) N_2O_2 + Cl_2 \xrightarrow{K_3} 2 \ NOCl$ (slow) The rate expression of the reaction will be :

When the concentration of a reactant,$A$,in a reaction $A \to \text{Products}$ is doubled,the rate of reaction increases seven times. The order of the reaction is between:

For the reaction $A + 2B \rightarrow \text{Product } (P)$,the rate law is given by $\frac{d[P]}{dt} = K[A]^2[B]$. If $[A]$ is taken in large excess,what will be the order of the reaction?

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The mechanism of reaction $2 \ NO + Cl_2 \rightarrow 2 \ NOCl$ is given as :
$(i) 2 NO \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} N_2 O _2$ (fast)
$(ii) N_2O_2 + Cl_2 \xrightarrow{K_3} 2 \ NOCl$ (slow)
The rate expression of the reaction will be :