A) \[18.39\text{ }kcal\text{ }mo{{l}^{-1}}\]
B) \[20\,kcal\,mo{{l}^{-1}}\]
C) \[16\,kcal\,mo{{l}^{-1}}\]
D) \[21.5\,kcal\,mo{{l}^{-1}}\]
Correct Answer: A
Solution :
[a] Let the initial concentration [a] = 100 Final concentration at \[298\text{ }K=100-10=90\] Final concentration at \[308\text{ }K=100-25=75\] Substituting the values in the 1st order rate reaction \[t=\frac{2.303}{{{k}_{298}}}\log \frac{100}{90}\] .... (i) \[t=\frac{2.303}{{{k}_{308}}}\log \frac{100}{75}\] ..... (ii) From (i) and (ii) \[\frac{{{k}_{308}}}{{{k}_{208}}}=2.73\] Substituting the value in the following relation \[{{E}_{a}}=\frac{2.303R\times {{T}_{1}}\times {{T}_{2}}}{{{T}_{2}}-{{T}_{1}}}\log \frac{{{k}_{2}}}{{{k}_{1}}}\] \[=\frac{2.303\times 8.314\times 298\times 308}{308-298}\log 2.73\] \[{{E}_{a}}=76.62\text{ }kJ\,mo{{l}^{-1}}=18.39\text{ }kcal\text{ }mo{{l}^{-1}}\]
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