The temperature at which $k_{1}=k_{2}$ will be
$10^{16} e^{-2000 / T}=10^{15} e^{-1000 / T}$
$\Rightarrow \frac{e^{-20001 T}}{e^{-1000 T}}=\frac{10^{15}}{10^{16}}$
$\Rightarrow e^{\frac{-1000}{T}}=10^{-1} \Rightarrow \log _{e} e^{\frac{-1000}{T}}=\log _{e} 10^{-1}$
$\Rightarrow 2.303 \log _{10} e^{\frac{-1000}{T}}=2.303 \times \log _{10} 10^{-1}$
$\Rightarrow \frac{-1000}{T} \times \log _{10} e=-1$ $\Rightarrow T=\frac{1000}{2.303} \mathrm{K}$
| પ્રયોગ | $\frac{[ X ]}{ mol \;L ^{-1}}$ | $\frac{[ Y ]}{ mol\; L ^{-1}}$ | $\frac{\text { Initial rate }}{ mol\; L ^{-1}\; min ^{-1}}$ |
| $I$ | $0.1$ | $0.1$ | $2 \times 10^{-3}$ |
| $II$ | $.2$ | $0.2$ | $4 \times 10^{-3}$ |
| $III$ | $0.4$ | $0.4$ | $M \times 10^{-3}$ |
| $IV$ | $0.1$ | $0.2$ | $2 \times 10^{-3}$ |
$M$ મૂલ્યનો સંખ્યાત્મક ગુણોત્તર $........$ છે. (નજીકનો પૂર્ણાંક)
| No | $[A_2]\, M$ | $[B_2]\, M$ | rate of reaction |
| $1.$ | $0.1\,M$ | $0.1\,M$ | $1.6 \times {10^{ - 4}}$ |
| $2.$ | $0.1\,M$ | $0.2\,M$ | $3.2 \times {10^{ - 4}}$ |
| $3.$ | $0.2\,M$ | $0.1\,M$ | $3.2 \times {10^{ - 4}}$ |
${{H}_{2}}+C{{l}_{2}}\xrightarrow{\text{Sunlight}}2HCl$