Medium
Consider the reaction $2 \ A + 2 \ B \rightarrow C + 2 \ D$. If the concentration of $A$ is doubled at constant $B$,the rate increases by a factor of $4$. If the concentration of $B$ is doubled at constant $A$,the rate is doubled. What is the rate law?
- A$r = k[A]^{2}[B]^{2}$
- B$r = k[A]^{4}[B]^{2}$
- C$r = k[A][B]^{2}$
- D$r = k[A]^{2}[B]$
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View SolutionTotal order of reaction $X + Y \rightarrow XY$ is $3$. The order of reaction with respect to $X$ is $2$. State the differential rate equation for the reaction.
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 rate of a reaction is given by $r = K[x][y] / [OH^-]$. If the concentration of $[OH^-]$ is increased,the order of the reaction will be ........
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