$\frac{{d\ln k}}{{dt}} = \frac{{\Delta {H^o}}}{{R{T^2}}}$
or $\ln {k_p} = - \frac{{\Delta {H^o}}}{{RT}} + I$.
Hence $(a)$ is correct $(b)$ is also correct as plot of $log\,(X)$ vs $time$ is linear.
$ (c) $ is wrong because $p \propto T$ at constant volume. $(d)$ is correct by Boyle’s law.
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$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}$ (Reaction 1$)$
$\mathrm{P} \rightarrow \mathrm{Q}$ (Reaction $2$)
The ratio of the half life of Reaction $1$ : Reaction $2$ is $5: 2$. If $t_1$ and $t_2$ represent the time taken to complete $2 / 3^{\text {dd }}$ and $4 / 5^{\text {dd }}$ of Reaction $1$ and
Reaction $2$, respectively, then the value of the ratio $\mathrm{t}_1: \mathrm{t}_2$ is . . . .$\times 10^{-1}$ (nearest integer).
[Given: $\log _{10}(3)=0.477$ and $\log _{10}(5)=0.699$ ]
| List-$I$ Species | List-$II$ Electronic distribution |
| $(A)$ $\mathrm{Cr}^{+2}$ | $(I)$ $3 \mathrm{~d}^8$ |
| $(B)$ $\mathrm{Mn}^{+}$ | $(II)$ $3 \mathrm{~d}^3 4 \mathrm{~s}^1$ |
| $(C)$ $\mathrm{Ni}^{+2}$ | $(III)$ $3\mathrm{~d}^4$ |
| $(D)$ $\mathrm{V}^{+}$ | $(IV)$ $3 \mathrm{~d}^5 4 \mathrm{~s}^1$ |
Choose the correct answer from the options given below:
$Z{n^{2 + }}\,(aq)\, + \,2e\, \rightleftharpoons \,Zn\,(s)\,;\, - \,0.762\,V$
$C{r^{3 + }}\,(aq)\, + \,3e\, \rightleftharpoons \,Cr(s)\,;\, - \,0.740\,\,V$
$2{H^ + }\,(aq)\, + \,2e\, \rightleftharpoons \,{H_2}(g)\,;\,\,\,0.00\,\,\,V$
$F{e^{3 + }}\,(aq)\, + \,e\, \rightleftharpoons \,F{e^{2 + }}(aq)\,;\,\,\,0.770\,\,\,V$
Which is the strongest reducing agent ?