A) \[80.77\,\,kJ\]
B) \[56.38\,\,kJ\]
C) \[24.67\,\,kJ\]
D) \[90.43\,\,kJ\]
Correct Answer: A
Solution :
\[\log \frac{{{k}_{2}}}{{{k}_{1}}}=\log 1000\] \[=\frac{{{E}_{1}}-{{E}_{2}}}{RT}\] \[=\frac{98000-{{E}_{2}}}{8.314\times 300}\] \[{{E}_{2}}\](catalyst)\[=80.77\,\,kJ\]You need to login to perform this action.
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