A) 12 kJ/mol
B) 831.4 kJ/mol
C) 100 kJ/mol
D) 88.57 kJ/mol
Correct Answer: C
Solution :
[c] \[{{e}^{-\frac{{{E}_{a}}}{RT}}}=3.8\times -\frac{{{10}^{-16}}}{100}\] \[-\frac{{{E}_{a}}}{RT}=\ln \,3.8\times {{10}^{-18}}\text{ }{{E}_{a}}=100kJ/mol\]You need to login to perform this action.
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