Question

For the two gaseous reactions, following data are given
$ A\xrightarrow{{}}B;{{k}_{1}}={{10}^{10}}{{e}^{-20,000/T}} $
$ C\xrightarrow{{}}D;{{k}_{2}}={{10}^{12}}{{e}^{-24,606/T}} $
the temperature at which $ {{k}_{1}} $ becomes equal to $ {{k}_{2}} $ is

• A. $400\, K$
• B. $1000\, K$
• C. $800\, K$
• D. $1500\, K$
• E. $500\, K$

Answer 1

Answer: (B)

Given $ {{k}_{1}}={{10}^{10}}{{e}^{-20,000/T}} $
$ {{k}_{2}}={{10}^{12}}{{e}^{-24,606/T}} $
$ {{k}_{1}}={{k}_{2}} $
$ {{10}^{10}}{{e}^{-20,000/T}}={{10}^{12}}{{e}^{-24,606/T}} $
$ {{e}^{\frac{-20,000}{T}+\frac{24,606}{T}}}={{10}^{2}} $
$ {{e}^{\frac{4,606}{T}}}={{10}^{2}} $
On taking log both sides
$ \frac{4606}{2.303T}=\log \,{{10}^{2}} $
$ 2\,\log 10\times T=\frac{4606}{2.303} $
$ T=\frac{4606}{2.303\times 2} $
$ =\frac{4606}{4.606}=1000\,K $