For the hypothetical reaction $2X + G \to Q + 2M$ , the rate expression is $\frac{{d\left[ Q \right]}}{{dt}} = k{\left[ X \right]^2}$ . Which of the following is the most likely mechanism ?
$2X + G \rightleftharpoons 2Q + R$ (fast)
$Q + R + G \to 2M$ (slow)
$X + G \rightleftharpoons Q + R$ (fast)
$R + X \to 2M$ (slow)
$X + X \rightleftharpoons {X_2}$ (fast)
${X_2} \to Q + T$ (slow)
$T + G \to 2M$ (fast)
$G + G \rightleftharpoons {G_2}$ (fast)
${G_2} + X \to Q + T$ (slow)
$T + X \to 2M$ (fast)
The rate of reaction between two reactants $A $ and $B$ decreases by a factor of $4$ if the concentration of reactant $B$ is doubled. The order of this reaction with respect to reactant $B$ is
Consider a reaction $\mathrm{aG}+\mathrm{bH} \rightarrow$ Products. When concentration of both the reactants $\mathrm{G}$ and $\mathrm{H}$ is doubled, the rate increases by eight times. However, when concentration of $\mathrm{G}$ is doubled keeping the concentration of $\mathrm{H}$ fixed, the rate is doubled. The overall order of the reaction is
The experimental data for decomposition of $N _{2} O _{5}$
$\left[2 N _{2} O _{5} \rightarrow 4 NO _{2}+ O _{2}\right]$
in gas phase at $318 \,K$ are given below:
$t/s$ | $0$ | $400$ | $800$ | $1200$ | $1600$ | $2000$ | $2400$ | $2800$ | $3200$ |
${10^2} \times \left[ {{N_2}{O_5}} \right]/mol\,\,{L^{ - 1}}$ | $1.63$ | $1.36$ | $1.14$ | $0.93$ | $0.78$ | $0.64$ | $0.53$ | $0.43$ | $0.35$ |
$(i)$ Plot $\left[ N _{2} O _{5}\right]$ against $t$
$(ii)$ Find the half-life period for the reaction.
$(iii)$ Draw a graph between $\log \left[ N _{2} O _{5}\right]$ and $t$
$(iv)$ What is the rate law $?$
$(v)$ Calculate the rate constant.
$(vi)$ Calculate the half-life period from $k$ and compare it with $(ii)$.
The conversion of molecules $X$ to $Y$ follows second order kinetics. If concentration of $X$ is increased to three times how will it affect the rate of formation of $Y ?$
For the reaction $3\,{A_{\,(g)\,}}\,\xrightarrow{K}\,{B_{(g)}}\, + \,\,{C_{(g)\,,}}K$ is ${10^{ - 14}}\,L/mol.\min .$ if $(A) = 0.5\,M$ then the value of $ - \frac{{d(A)}}{{dt}}$ (in $M / sec$ ) is.