The rate constant for a second order reaction is $8 \times {10^{ - 5}}\,{M^{ - 1}}\,mi{n^{ - 1}}$. How long will it take a $ 1\,M $ solution to be reduced to $0.5\, M$

  • A

    $8 \times {10^{ - 5}}\,\min$

  • B

    $8.665 \times {10^3}\,\min$

  • C

    $4 \times {10^{ - 5}}\,\min$

  • D

    $1.25 \times {10^4}\,\min$

Similar Questions

Write unit of rate constant of following reaction :

$1.$ fourth order

$2.$ third order

For reaction :

$2NO_2(g) + O_3(g) \to N_2O_5(g) + O_2(g)$

rate law is $R = K\, [NO_2]' [O_3]'$.

Which of these possible reaction mechanisms is consistent with the rate law?

Mechanism $I :$

$NO_2(g) + O_3(g) \to NO_3(g) + O_2(g)$ (slow)

$NO_3(g) + NO_2(g) \to N_2O_5(g)$ (fast)

Mechanism $II :$

$O_3(g)  \rightleftharpoons  O_2(g) + [O]$ (fast)

$NO_2(g) + [O] \to NO_3$ (slow)

$NO_3(g) + NO_2(g) \to  N_2O_5$ (fast)

The rate of certain reaction depends on concentration according to the equation $\frac{{ - dc}}{{dt}}\, = \,\frac{{{K_1}C}}{{1 + {K_2}C}},$ what is the order, when concentration $(c)$ is very-very high

For the reaction $A + B \to C$, it is found that doubling the concentration of $A$ increases the rate by $4$ times, and doubling the concentration of $B$ doubles the reaction rate. What is the overal order of the reaction.

For the reaction $A + B \to $ products, what will be the order of reaction with respect to $A$ and $B$ ?

   Exp.    $[A]\,(mol\,L^{-1})$   $[B]\,(mol\,L^{-1})$   Initial rate    $(mol\,L^{-1}\,s^{-1})$ 
   $1.$  $2.5\times 10^{-4}$  $3\times 10^{-5}$  $5\times 10^{-4}$
   $2.$   $5\times 10^{-4}$  $6\times 10^{-5}$  $4\times 10^{-3}$
   $3.$   $1\times 10^{-3}$  $6\times 10^{-5}$  $1.6\times 10^{-2}$