Saved Bookmarks
| 1. |
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 theabove 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? |
|
Answer» `10.14 kcal//mol` |
|