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How do I use the half-life formula of a first order reaction in 2nd PUC Chemical Kinetics problems?

GradeHunt_TeamGradeHunt Team · · Karnataka Board 12th

2nd PUC Chemistry, Chemical Kinetics. I know t1/2 = 0.693/k for first order, but in problems they ask time for 90% completion or how much is left after some time, and I mess up the log part. How is half-life different for zero order?
chemical kineticshalf lifefirst order reaction2nd puc chemistry

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For a first order reaction the integrated rate equation is: k = (2.303 / t) log ([R]0 / [R]) [R]0 is the initial concentration and [R] is the concentration after time t. Half-life: at t = t1/2, [R] = [R]0 / 2, so [R]0/[R] = 2. k = (2.303 / t1/2) log 2 = (2.303 x 0.3010) / t1/2 t1/2 = 0.693 / k The key point: for a first order reaction, half-life does not depend on the initial concentration. Every half-life cuts the amount by half, whatever you start with. Worked example: k = 0.0693 min^-1. (a) Half-life: t1/2 = 0.693 / 0.0693 = 10 min (b) Fraction left after 30 min: 30 min = 3 half-lives, so the fraction left = 1/2 x 1/2 x 1/2 = 1/8, which is 12.5%. If the time is not a whole number of half-lives, use the log form. After 15 min: log([R]0/[R]) = kt/2.303 = (0.0693 x 15)/2.303 = 0.451. So [R]0/[R] = antilog 0.451 = 2.83, and [R] = [R]0/2.83, which is about 35% left. (c) Time for 90% completion: If 90% has reacted, 10% is left. So [R]0/[R] = 100/10 = 10, and log 10 = 1. t = (2.303 / k) x 1 = 2.303 / 0.0693 = 33.2 min (approx.) Compare with zero order: t1/2 = [R]0 / 2k Here half-life is directly proportional to the initial concentration. Double the starting concentration and the half-life doubles. Common slips: 1. Writing [R]/[R]0 inside the log. It is always initial over remaining, so the log is positive. 2. Using 'percentage completed' directly. Convert to 'percentage left' first: 75% completion means 25% left, so the ratio is 4. 3. Mixing units. If k is in s^-1, the time comes out in seconds. Useful shortcut: 75% completion of a first order reaction takes exactly 2 half-lives, and 87.5% takes 3.
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