$A + B \rightarrow \text{product}$. The rate of this reaction is given by $v = K[A]^2[B]^0$. What is the change in the rate of reaction when the concentration of $A$ is doubled and the concentration of $B$ is doubled?
- A$4$ times
- B$2$ times
- C$8$ times
- D$1/4$ times
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Similar Questions
$A$ reaction is first order in terms of $A$ and second order in terms of $B$. What will be the rate of reaction,if the concentration of $B$ is increased two times?
The rate of the reaction $CCl_3CHO + NO \to CHCl_3 + NO + CO$ is given by $Rate = K [CCl_3CHO] [NO]$. If concentration is expressed in $mol \ L^{-1}$,the units of $K$ are:
Easy
View SolutionFor the reaction,$PCl_5 \longrightarrow PCl_3 + Cl_2$,the rate and rate constant are $1.02 \times 10^{-4} \ mol \ L^{-1} \ s^{-1}$ and $3.4 \times 10^{-5} \ s^{-1}$ respectively at a given instant. The molar concentration of $PCl_5$ at that instant is
MediumKCET 2024
View SolutionFor 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:
For the reaction $2A + B \rightarrow \text{Product}$,the following mechanism is given. What is the order of the reaction?
$2A \rightleftharpoons A_2$ (fast)
$A_2 + B \rightarrow P$ (slow)
$2A \rightleftharpoons A_2$ (fast)
$A_2 + B \rightarrow P$ (slow)
Medium
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