The rate of the reaction,$CH_3COOC_2H_5 + NaOH \longrightarrow CH_3COONa + C_2H_5OH$ is given by the equation,$\text{rate} = k[CH_3COOC_2H_5][NaOH]$. If concentration is expressed in $mol \ L^{-1}$,the unit of $k$ is
- A$mol \ L^{-1} \ s^{-1}$
- B$L \ mol^{-1} \ s^{-1}$
- C$s^{-1}$
- D$mol^{-2} \ L^2 \ s^{-1}$
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The rate of the reaction becomes twice when the concentration of reactant becomes $8$ times,then the order of the reaction is:
Medium
View SolutionWhat is the order and molecularity respectively for the elementary reaction given below?
$O_{3(g)} + O_{(g)} \rightarrow 2O_{2(g)}$ if $r = k[O_3][O]$
$O_{3(g)} + O_{(g)} \rightarrow 2O_{2(g)}$ if $r = k[O_3][O]$
For a reaction,$A + B \longrightarrow$ Product; the rate law is given by $r = k[A]^2[B]^{1/2}$. What is the order of the reaction?
Medium
View SolutionThe rate law for the decomposition of hydrogen iodide is $-\frac{d[HI]}{dt}=k[HI]^2$. The units of rate constant $k$ are
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 $(mol/L \cdot s)$ |
| $0.05$ | $0.05$ | $1.2 \times 10^{-3}$ |
| $0.10$ | $0.05$ | $2.4 \times 10^{-3}$ |
| $0.05$ | $0.10$ | $1.2 \times 10^{-3}$ |
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