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 $0.5\, M$
$8 \times {10^{ - 5}}\,\min$
$8.665 \times {10^3}\,\min$
$4 \times {10^{ - 5}}\,\min$
$1.25 \times {10^4}\,\min$
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 $(mol\,L^{-1}\,s^{-1})$ |
$1.$ | $2.5\times 10^{-4}$ | $3\times 10^{-5}$ | $5\times 10^{-4}$ |
$2.$ | $5\times 10^{-4}$ | $6\times 10^{-5}$ | $4\times 10^{-3}$ |
$3.$ | $1\times 10^{-3}$ | $6\times 10^{-5}$ | $1.6\times 10^{-2}$ |