Q38 / 170 · Manipal (MET) 2015 · Chemistry

For the two gaseous reactions, following data $\begin{aligned} & A \longrightarrow B ; k_1=10^{10} e^{-20 ; 000 / T} \\ & C \longrightarrow D ; k_2=10^{12} e^{-24,606 / T}\end{aligned}$ the temperature at which $k_1$ becomes equal to $k_2$ is

  • A
    400 K
  • Correct answer
    B
    1000 K
  • C
    800 K
  • D
    1500 K

Solution

Given $_{\star} k_1=10^{10} e^{-20,000 / T}$ $\begin{aligned} & k_2=10^{12} e^{-24,606 / T} \\ & k_1=k_2 \Rightarrow 10^{10} e^{-20,000 / T}=10^{12} e^{-24,606 / T}\end{aligned}$ $e^{-\frac{20,000}{T}+\frac{24606}{T}}=10^2$ $e^{\frac{4,606}{T}}=10^2$ On taking log both sides, $\begin{gathered}\frac{4.606}{2.303 T}=\log 10^2 \\ 2 \log 10 \times T=\frac{4606}{2.303} \\ T=\frac{4606}{2.303 \times 2}=\frac{4606}{4.606}=1000 \mathrm{~K}\end{gathered}$

Hence, the correct answer is option: 'B'.

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