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 = 0.69)

Question

Dr. Abubacker · Accepted Answer

[Solution from source] k = (2.303/60) log(100/40) = (2.303/60)(0.69 − 0.30) ≈ 0.0150 min⁻¹. t½ = 0.693/0.0150 ≈ 46 min, closest to 45 minutes. 50 minutes assumes a linear (zero-order) fall. Answer (1).

Q. 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 = 0.69)

Explanation

Reviewed by Dr. Abubacker

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