If 60% of a first order reaction was completed in 60 minutes, 50% of the same reaction would be completed in approximately (log 4 = 0.60, log 5 = 0.70)
- 45 minutes
- 60 minutes
- 40 minutes
- 50 minutes
Answer
(1) 45 minutes
Solution
For a first order reaction, k = 2.303t [R]0[R]
60% completed ⇒ [R] = 40 when [R]0 = 100
k = 2.30360 10040
= 2.30360( 10 - 4) = 2.303 0.4060
For 50% completion, t1/2 = 2.303k 2, with 2 = 12 4 = 0.30
⇒ t1/2 = 2.303 0.30 602.303 0.40
= 45 minutes
Answer: (1)