Medium
For the reaction $2 \ NOBr_{(g)} \rightarrow 2 \ NO_{(g)} + Br_{2_{(g)}}$,the rate law is $r = k[NOBr]^{2}$. If the rate constant is $1.62 \ M^{-1} \ s^{-1}$ and the concentration of $NOBr$ is $2.00 \times 10^{-3} \ M$,what is the rate of reaction?
- A$6.48 \times 10^{-6} \ M \ s^{-1}$
- B$4.05 \times 10^{-5} \ M \ s^{-1}$
- C$2.46 \times 10^{-6} \ M \ s^{-1}$
- D$5.24 \times 10^{-6} \ M \ s^{-1}$
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For the gaseous reaction,$N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2$,the rate can be expressed as:
$-\frac{d[N_2O_5]}{dt} = K_1[N_2O_5]$
$+\frac{d[NO_2]}{dt} = K_2[N_2O_5]$
$+\frac{d[O_2]}{dt} = K_3[N_2O_5]$
The correct relation between $K_1, K_2$ and $K_3$ is:
$-\frac{d[N_2O_5]}{dt} = K_1[N_2O_5]$
$+\frac{d[NO_2]}{dt} = K_2[N_2O_5]$
$+\frac{d[O_2]}{dt} = K_3[N_2O_5]$
The correct relation between $K_1, K_2$ and $K_3$ is:
The reaction $2NO_{(g)} + O_{2(g)} \rightleftharpoons 2NO_{2(g)}$ is of first order. If the volume of the reaction vessel is reduced to $1/3$ of its original volume,the rate of reaction would be .......... times.
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View SolutionFor a reaction $A \to B$,the rate of reaction quadrupled when the concentration of $A$ is doubled. The rate expression of the reaction is $r = K[A]^n$. The value of $n$ is
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View SolutionThe reaction between $A$ and $B$ is first order with respect to $A$ and zero order with respect to $B$. Fill in the blanks in the following table:
Experiment $[A] / mol \, L^{-1}$ $[B] / mol \, L^{-1}$ Initial rate / $mol \, L^{-1} \, min^{-1}$ $I$ $0.1$ $0.1$ $2.0 \times 10^{-2}$ $II$ $-$ $0.2$ $4.0 \times 10^{-2}$ $III$ $0.4$ $0.4$ $-$ $IV$ $-$ $0.2$ $2.0 \times 10^{-2}$
| Experiment | $[A] / mol \, L^{-1}$ | $[B] / mol \, L^{-1}$ | Initial rate / $mol \, L^{-1} \, min^{-1}$ |
|---|---|---|---|
| $I$ | $0.1$ | $0.1$ | $2.0 \times 10^{-2}$ |
| $II$ | $-$ | $0.2$ | $4.0 \times 10^{-2}$ |
| $III$ | $0.4$ | $0.4$ | $-$ |
| $IV$ | $-$ | $0.2$ | $2.0 \times 10^{-2}$ |
Difficult
View SolutionFor the reaction $2A + B \to C$,the values of initial rate at different reactant concentrations are given in the table below: The rate law for the reaction is
$[A] \ (mol \ L^{-1})$ $[B] \ (mol \ L^{-1})$ Initial Rate $(mol \ L^{-1} \ s^{-1})$ $0.05$ $0.05$ $0.045$ $0.10$ $0.05$ $0.090$ $0.20$ $0.10$ $0.72$
| $[A] \ (mol \ L^{-1})$ | $[B] \ (mol \ L^{-1})$ | Initial Rate $(mol \ L^{-1} \ s^{-1})$ |
| $0.05$ | $0.05$ | $0.045$ |
| $0.10$ | $0.05$ | $0.090$ |
| $0.20$ | $0.10$ | $0.72$ |
DifficultJEE MAIN 2019
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