Medium
Rate law for a reaction between $A$ and $B$ is given by $R=k[A]^{n}[B]^{m}$. If concentration of $A$ is doubled and concentration of $B$ is halved from their initial value,the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{r_2}{r_1}\right)$ is
- A$2^{(n-m)}$
- B$(n-m)$
- C$(m+n)$
- D$\frac{1}{2^{m+n}}$
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Similar Questions
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:
In a reaction,if the concentration of reactant $A$ is tripled,the rate of reaction becomes twenty-seven times. What is the order of the reaction?
Easy
View SolutionIf the half-life of a reaction is halved when the initial concentration of the reactant is doubled,what is the order of the reaction?
Medium
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 a reaction $r = k[A][B]^2$,if the concentration of $A$ is doubled,the rate of reaction:
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