The rate of the reaction becomes twice when the concentration of reactant becomes $8$ times then the order of the reaction is
The rate of the reaction becomes twice when the concentration of reactant becomes $8$ times then the order of the reaction is
- A
$3$
- B
$\frac{1}{2}$
- C
$\frac{1}{3}$
- D
$1$
Similar Questions
For the non - stoichimetre reaction $2A + B \rightarrow C + D,$ the following kinetic data were obtained in three separate experiments, all at $298\, K.$
Initial Concentration
$(A)$
Initial Concentration
$(A)$
Initial rate of formation of $C$
$(mol\,L^{-1}\,s^{-1})$
$0.1\,M$
$0.1\,M$
$1.2\times 10^{-3}$
$0.1\,M$
$0.2\,M$
$1.2\times 10^{-3}$
$0.2\,M$
$0.1\,M$
$2.4 \times 10^{-3}$
The rate law for the formation of $C$ is :
For the non - stoichimetre reaction $2A + B \rightarrow C + D,$ the following kinetic data were obtained in three separate experiments, all at $298\, K.$
|
Initial Concentration $(A)$ |
Initial Concentration $(A)$ |
Initial rate of formation of $C$ $(mol\,L^{-1}\,s^{-1})$ |
| $0.1\,M$ | $0.1\,M$ | $1.2\times 10^{-3}$ |
| $0.1\,M$ | $0.2\,M$ | $1.2\times 10^{-3}$ |
| $0.2\,M$ | $0.1\,M$ | $2.4 \times 10^{-3}$ |
The rate law for the formation of $C$ is :
- [JEE MAIN 2014]
Write general equation of reaction and explain - what is order of reaction ? Which is its value ?
Write general equation of reaction and explain - what is order of reaction ? Which is its value ?
The data for the reaction $A + B \to C$ is
Exp
$[A]_0$
$[B]_0$
initial rate
$1$
$0.012$
$0.035$
$0.10$
$2$
$0.024$
$0.035$
$0.80$
$3$
$0.012$
$0.070$
$0.10$
$4$
$0.024$
$0.070$
$0.80$
The data for the reaction $A + B \to C$ is
| Exp | $[A]_0$ | $[B]_0$ | initial rate |
| $1$ | $0.012$ | $0.035$ | $0.10$ |
| $2$ | $0.024$ | $0.035$ | $0.80$ |
| $3$ | $0.012$ | $0.070$ | $0.10$ |
| $4$ | $0.024$ | $0.070$ | $0.80$ |
For the first order decompsition reaction of $N_2O_5$, it is found that -
$(a)$ $2N_2O_5\rightarrow\,\,4NO_2(g)+O_2(g)-\frac{d[N_2O_5]}{dt}=k[N_2O_5]$
$(a)$ $N_2O_5\rightarrow\,\,2NO_2(g)+1/2\,\,O_2(g)-\frac{d[N_2O_5]}{dt}=k'[N_2O_5]$
which of the following is true ?
For the first order decompsition reaction of $N_2O_5$, it is found that -
$(a)$ $2N_2O_5\rightarrow\,\,4NO_2(g)+O_2(g)-\frac{d[N_2O_5]}{dt}=k[N_2O_5]$
$(a)$ $N_2O_5\rightarrow\,\,2NO_2(g)+1/2\,\,O_2(g)-\frac{d[N_2O_5]}{dt}=k'[N_2O_5]$
which of the following is true ?
The rate constant for the reaction, $2{N_2}{O_5} \to 4N{O_2}$ $ + {O_2}$ is $3 \times {10^{ - 5}}{\sec ^{ - 1}}$. If the rate is $2.40 \times {10^{ - 5}}\,mol\,\,litr{e^{{\rm{ - 1}}}}{\sec ^{ - 1}}$. Then the concentration of ${N_2}{O_5}$ (in mol litre $^{-1}$) is
The rate constant for the reaction, $2{N_2}{O_5} \to 4N{O_2}$ $ + {O_2}$ is $3 \times {10^{ - 5}}{\sec ^{ - 1}}$. If the rate is $2.40 \times {10^{ - 5}}\,mol\,\,litr{e^{{\rm{ - 1}}}}{\sec ^{ - 1}}$. Then the concentration of ${N_2}{O_5}$ (in mol litre $^{-1}$) is
- [IIT 2000]