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3. Chemical kinetics — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistry3. Chemical kinetics1 Mark
MCQ
The rate constants $k_1$ and $k_2$ for two differentre actions are $10^{16} \cdot e^{-2000/T}$ and $ 10^{15} \cdot e^{-1000/T},$ respectively. The temperature at which $k_1 = k_2$ is
  • A
    $2000\,K$
  • ✓
    $\frac{1000}{2.303} \,K$
  • C
    $1000\,K$
  • D
    $\frac{2000}{2.303} \,K$
✓

Answer

Correct option: B.
$\frac{1000}{2.303} \,K$
b
$k_{1}=10^{16} e^{-2000 T}, k_{2}=10^{15} e^{-1000 / T}$

The temperature at which $k_{1}=k_{2}$ will be

$10^{16} e^{-2000 / T}=10^{15} e^{-1000 / T}$

$\Rightarrow \frac{e^{-20001 T}}{e^{-1000 T}}=\frac{10^{15}}{10^{16}}$

$\Rightarrow e^{\frac{-1000}{T}}=10^{-1} \Rightarrow \log _{e} e^{\frac{-1000}{T}}=\log _{e} 10^{-1}$

$\Rightarrow 2.303 \log _{10} e^{\frac{-1000}{T}}=2.303 \times \log _{10} 10^{-1}$

$\Rightarrow \frac{-1000}{T} \times \log _{10} e=-1$ $\Rightarrow T=\frac{1000}{2.303} \mathrm{K}$

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