A) \[2.303\times 2\,cal,5\]
B) \[\frac{2}{3.303}cal,{{e}^{5}}\]
C) 2 cal, 5
D) none of these
Correct Answer: C
Solution :
\[K=A{{e}^{\frac{Ea}{RT}}}\] |
\[Log\,\,K=\log \,A-\frac{Ea}{2.303R}\times \frac{1}{T}\] |
Slope=\[\tan \theta =\left[ -\frac{1}{2.303} \right]=-\frac{{{E}_{a}}}{2.303R}\] |
So, \[Ea=R=2cal/mol\] |
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