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A 3 mol mixture of substances A and B (mole fraction of A = 2/3) undergoes parallel first order reactions to give product C: A -> C with t1/2 = 2 hr, and B -> C with t1/2 = 4 hr. At what time will 2 mol of C have formed?
- 2 hr
- 4 hr
- 6 hr
- 8 hr
Correct answer: 4 hr
Solution
At 4 hr A (t1/2 = 2 hr) has completed 2 half-lives leaving 0.5 mol (1.5 mol -> C) and B (t1/2 = 4 hr) leaves 0.5 mol (0.5 mol -> C), giving total C = 2 mol.
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