An analyst starts a first order chemica reaction at 8.00 A.M. in the morning at the laboratory temperature of `27^(@)` C At 1.00 P.M. he discovered that only 10% of the reaction was complete by that time To speed-up the reaction he increased the temperature to `127^(@)` C At 4.00 P.M. he found that only 50% of the reaction was complete Any how he did not want to stay in laboratory beyond 5.00 P.M. but he could not leave the laboratory until the reaction was 90% complete Fortunately he found a suitable catayst adding which at 4.00 P.M. at `127^(@)` C he could meet the target of 5.00 P.M. and 90% Answer the following questions based on the above observation (Use `l n 10/9 = 0.1 l n 9/5=0.6, l n10=2.3 l n5=1.6,l n8=2`)
What was the activation energy of the original pathway?

Options

A. `10.14 kcal//mol`
B. `5.52 kcal//mol`
C. `2.64 kcal//mol`
D. `7.92 kcal//mol`

Answer

B