The rate constant k, for the reaction ${N_2}{O_5}(g) \to $ $2N{O_2}(g) + \frac{1}{2}{0_2}(g)$ is $2.3 \times {10^{ - 2}}\,{s^{ - 1}}$. Which equation given below describes the change of $[{N_2}{O_5}]$ with time? ${[{N_2}{O_5}]_0}$ and ${[{N_2}{O_5}]_t}$ correspond to concentration of ${N_2}{O_5}$ initially and at time $t$.
${[{N_2}{O_5}]_t} = {[{N_2}{O_5}]_0} + kt$
${[{N_2}{O_5}]_0} = {[{N_2}{O_5}]_t}{e^{kt}}$
${\log _{10}}{[{N_2}{O_5}]_t} = {\log _{10}}{[{N_2}{O_5}]_0} - kt$
${\rm{ln}}\frac{{{{{\rm{[}}{{\rm{N}}_{\rm{2}}}{O_5}]}_0}}}{{{{{\rm{[}}{{\rm{N}}_{\rm{2}}}{O_5}]}_t}}} = kt$
Consider the reaction :
$Cl_2(aq) + H_2S(aq) \to S(s) + 2H^+(aq) + 2Cl^-(aq)$
The rate equation for this reaction is rate $= k[Cl_2][H_2S]$ Which of these mechanisms is/are consistent with this rate equation ?
$A.\,C{l_2} + {H_2}S \to {H^ + } + C{l^ - } + C{l^ + } + H{S^- }$ (slow)
$C{l^ + } + H{S^ - } \to {H^ + } + C{l^ - } + {S}$ (fast)
$B.\, H_2S \Leftrightarrow H^+ + HS^-$ (fast equilibrium)
$Cl_2 + HS^-\to 2Cl^-+ H^+ + S$ (slow)
The following results were obtained during kinetic studies of the reaction $2A+B$ $\to$ products
Experiment |
$[A]$ (in $mol\, L^{-1})$ |
$[B]$ (in $mol\, L^{-1})$ |
Initial rate of reaction (in $mol\, L^{-1}\,min^{-1})$ |
$I$ | $0.10$ | $0.20$ | $6.93 \times {10^{ - 3}}$ |
$II$ | $0.10$ | $0.25$ | $6.93 \times {10^{ - 3}}$ |
$III$ | $0.20$ | $0.30$ | $1.386 \times {10^{ - 2}}$ |
The time(in minutes) required to consume half of $A$ is
For a given reaction $t_{1/2} = \frac{1}{k.a}$ the order of reaction will be
The mechanism of the reaction $A + 2B \to D$ is
$2B\xrightarrow{k}{B_2}$ [Slow]
${B_2} + A \to D$ [Fast]
The rate law expression, order with respect to $A$, order with respect to $'B'$ and overall order are respectively
The rate of the reaction, $2NO + Cl_2 \rightarrow 2NOCl$ is given by the rate equation rate $= k[NO]^2[Cl_2].$ The value of the rate constant can be increased by