KCET · Chemistry · Chemical Kinetics
The rate constant \(k_1\) and \(k_2\) for two different reactions are \(10^{16} \times e^{-2000 / T}\) and \(10^{15} \times e^{-1000 / T}\) respectively. The temperature at which \(k_1=k_2\) is
- A \(\frac{2000}{2.303} \mathrm{~K}\)
- B \(2000 \mathrm{~K}\)
- C \(\frac{1000}{2.303} \mathrm{~K}\)
- D \(1000 \mathrm{~K}\)
Answer & Solution
Correct Answer
(C) \(\frac{1000}{2.303} \mathrm{~K}\)
Step-by-step Solution
Detailed explanation
\(K_1=10^{16} \times e^{-\frac{2000}{T}}\)
\(K_2=10^{15} \times e^{-\frac{1000}{T}}\)
Now, \(K_1=K_2\)
\(\begin{aligned} 10^{16} \times e^{\frac{-2000}{T}} & =10^{15} \times e^{\frac{-1000}{T}} \\ 10 & =e^{\frac{1000}{T}}\end{aligned}\)
Taking \(\ln\) on both sides,
\ln 10=\frac{1000}{T} \ln e
\(2.303 \log 10=\frac{1000}{T} \Rightarrow T=\frac{1000}{2.303} \mathrm{~K}\)
\(K_2=10^{15} \times e^{-\frac{1000}{T}}\)
Now, \(K_1=K_2\)
\(\begin{aligned} 10^{16} \times e^{\frac{-2000}{T}} & =10^{15} \times e^{\frac{-1000}{T}} \\ 10 & =e^{\frac{1000}{T}}\end{aligned}\)
Taking \(\ln\) on both sides,
\ln 10=\frac{1000}{T} \ln e
\(2.303 \log 10=\frac{1000}{T} \Rightarrow T=\frac{1000}{2.303} \mathrm{~K}\)
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