For the following parallel chain reaction. What will be that value of overall half-life of $A$ in minutes ?
Given that $\left[ {\frac{{{{\left[ B \right]}_t}}}{{{{[C]}_t}}} = \frac{{16}}{9}} \right]$
$A\,\xrightarrow{{{K_1}\, = \,2\, \times \,{{10}^{^{ - 3}\,}}{S^{ - 1}}}}4B$
$A\to C$
For the following parallel chain reaction. What will be that value of overall half-life of $A$ in minutes ?
Given that $\left[ {\frac{{{{\left[ B \right]}_t}}}{{{{[C]}_t}}} = \frac{{16}}{9}} \right]$
$A\,\xrightarrow{{{K_1}\, = \,2\, \times \,{{10}^{^{ - 3}\,}}{S^{ - 1}}}}4B$
$A\to C$
- A
$693$
- B
$\frac{693}{210}$
- C
$\frac{693}{30}$
- D
$\frac{693}{7}$
Similar Questions
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]
Following is the rate constant of reaction what is the overall order of reaction ?
$(a)$ $6.66 \times 10^{-3} \,s ^{-1}$
$(b)$ $4.5 \times 10^{-2} \,mol ^{-1} \,L \,s ^{-1}$
Following is the rate constant of reaction what is the overall order of reaction ?
$(a)$ $6.66 \times 10^{-3} \,s ^{-1}$
$(b)$ $4.5 \times 10^{-2} \,mol ^{-1} \,L \,s ^{-1}$
The order of a reaction which has the rate expression $\frac{{dc}}{{dt}} = K{[E]^{3/2}}{[D]^{3/2}}$ is
The order of a reaction which has the rate expression $\frac{{dc}}{{dt}} = K{[E]^{3/2}}{[D]^{3/2}}$ is
The following data was obtained for chemical reaction given below at $975\, \mathrm{~K}$.
$2 \mathrm{NO}_{(\mathrm{g})}+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{N}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}$
$[NO]$
$\mathrm{mol} \mathrm{L}^{-1}$
${H}_{2}$
$\mathrm{mol} \mathrm{L}^{-1}$
Rate
$\mathrm{mol}L^{-1}$ $s^{-1}$
$(A)$
$8 \times 10^{-5}$
$8 \times 10^{-5}$
$7 \times 10^{-9}$
$(B)$
$24 \times 10^{-5}$
$8 \times 10^{-5}$
$2.1 \times 10^{-8}$
$(C)$
$24 \times 10^{-5}$
$32 \times 10^{-5}$
$8.4 \times 10^{-8}$
The order of the reaction with respect to $\mathrm{NO}$ is ..... .
The following data was obtained for chemical reaction given below at $975\, \mathrm{~K}$.
$2 \mathrm{NO}_{(\mathrm{g})}+2 \mathrm{H}_{2(\mathrm{~g})} \rightarrow \mathrm{N}_{2(\mathrm{~g})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{g})}$
|
$[NO]$ $\mathrm{mol} \mathrm{L}^{-1}$ |
${H}_{2}$ $\mathrm{mol} \mathrm{L}^{-1}$ |
Rate $\mathrm{mol}L^{-1}$ $s^{-1}$ |
|
| $(A)$ | $8 \times 10^{-5}$ | $8 \times 10^{-5}$ | $7 \times 10^{-9}$ |
| $(B)$ | $24 \times 10^{-5}$ | $8 \times 10^{-5}$ | $2.1 \times 10^{-8}$ |
| $(C)$ | $24 \times 10^{-5}$ | $32 \times 10^{-5}$ | $8.4 \times 10^{-8}$ |
The order of the reaction with respect to $\mathrm{NO}$ is ..... .
- [JEE MAIN 2021]
Reaction $2A + B \to$ product, rate law is $\frac{{ - d[A]}}{{dt}}\, = \,K[A].$ At a time when $t\, = \,\frac{{{t_{1/2}}}}{{\ln\,2}},$ concentration of the reactant is
Reaction $2A + B \to$ product, rate law is $\frac{{ - d[A]}}{{dt}}\, = \,K[A].$ At a time when $t\, = \,\frac{{{t_{1/2}}}}{{\ln\,2}},$ concentration of the reactant is