Medium
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:
- A$K_1 = 2K_2 = 4K_3$
- B$2K_1 = K_2 = 4K_3$
- C$2K_1 = 3K_2 = 4K_3$
- D$4K_1 = 2K_2 = K_3$
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Similar Questions
The reaction between $X$ and $Y$ is first order with respect to $X$ and zero order with respect to $Y$.
$Experiment$ $[X] / (mol \ L^{-1})$ $[Y] / (mol \ L^{-1})$ $\text{Initial rate} / (mol \ L^{-1} \ min^{-1})$ $I$ $0.1$ $0.1$ $2 \times 10^{-3}$ $II$ $0.2$ $0.2$ $4 \times 10^{-3}$ $III$ $0.4$ $0.4$ $M \times 10^{-3}$ $IV$ $0.1$ $0.2$ $2 \times 10^{-3}$
Examine the data of the table and calculate the ratio of the numerical value of $M$ to $0.2$.
| $Experiment$ | $[X] / (mol \ L^{-1})$ | $[Y] / (mol \ L^{-1})$ | $\text{Initial rate} / (mol \ L^{-1} \ min^{-1})$ |
| $I$ | $0.1$ | $0.1$ | $2 \times 10^{-3}$ |
| $II$ | $0.2$ | $0.2$ | $4 \times 10^{-3}$ |
| $III$ | $0.4$ | $0.4$ | $M \times 10^{-3}$ |
| $IV$ | $0.1$ | $0.2$ | $2 \times 10^{-3}$ |
Examine the data of the table and calculate the ratio of the numerical value of $M$ to $0.2$.
The value of the rate constant of a pseudo first order reaction:
What is the time taken for the $3^{rd}$ half-life of a second-order decomposition reaction,given that its first half-life is $20 \ s$ (in $s$)?
Medium
View Solution$A$ reaction is of second order with respect to carbon monoxide. If the concentration of carbon monoxide is doubled,keeping everything else constant,the rate of the reaction will...
Easy
View SolutionThe possible mechanism for the reaction $2NO + Br_2 \to 2NOBr$ is:
$NO + Br_2 \rightleftharpoons NOBr_2$ (Fast)
$NOBr_2 + NO \to 2NOBr$ (Slow)
The rate law expression is:
$NO + Br_2 \rightleftharpoons NOBr_2$ (Fast)
$NOBr_2 + NO \to 2NOBr$ (Slow)
The rate law expression is:
Medium
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