$A + B \rightarrow \text{product}$. 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
Which of the following rate laws has an overall order of $0.5$ for a reaction involving substances $x$,$y$,and $z$?
MediumAIIMS 1983
View Solution$A$ chemical reaction takes place in two steps as follows:
$(i) \ NO_2Cl_{(g)} \xrightarrow{K_1} NO_{2(g)} + Cl_{(g)}$
$(ii) \ NO_2Cl_{(g)} + Cl_{(g)} \xrightarrow{K_2} NO_{2(g)} + Cl_{2(g)}$
Identify the reaction intermediate.
$(i) \ NO_2Cl_{(g)} \xrightarrow{K_1} NO_{2(g)} + Cl_{(g)}$
$(ii) \ NO_2Cl_{(g)} + Cl_{(g)} \xrightarrow{K_2} NO_{2(g)} + Cl_{2(g)}$
Identify the reaction intermediate.
Calculate the order of the reaction with respect to $A$ and $B$ based on the following data:
$[A] \ (mol/L)$ $[B] \ (mol/L)$ Rate $(mol/L \cdot s)$ $0.05$ $0.05$ $1.2 \times 10^{-3}$ $0.10$ $0.05$ $2.4 \times 10^{-3}$ $0.05$ $0.10$ $1.2 \times 10^{-3}$
| $[A] \ (mol/L)$ | $[B] \ (mol/L)$ | Rate $(mol/L \cdot s)$ |
| $0.05$ | $0.05$ | $1.2 \times 10^{-3}$ |
| $0.10$ | $0.05$ | $2.4 \times 10^{-3}$ |
| $0.05$ | $0.10$ | $1.2 \times 10^{-3}$ |
Easy
View SolutionThe reaction $2FeCl_3 + SnCl_2 \to 2FeCl_2 + SnCl_4$ is an example of
MediumAIPMT 1996
View SolutionWhen the temperature increases from $300 \, K$ to $310 \, K$,the rate of reaction increases by $2.3$ times. If the rate constant at $300 \, K$ is $x$,then the rate constant at $310 \, K$ will be equal to .......
Medium
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