A) 400 K
B) 1000 K
C) 800 K
D) 1500 K
E) 500 K
Correct Answer: B
Solution :
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 \[\frac{4606}{2.303T}=\log \,{{10}^{2}}\] \[2\,\log 10\times T=\frac{4606}{2.303}\] \[T=\frac{4606}{2.303\times 2}\] \[=\frac{4606}{4.606}=1000\,K\]You need to login to perform this action.
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