Question 52 - The half-life for first order reaction could be described in the differential equation dC / dt = -kC where k is a constant, C is concentration and t is time. (a) Find the equation of C as a function of t. (b) Find the half life for such reaction or the time required to reduce 50 % of the initial concentration, where k = 0.139 per minute. (c) When the initial concentration Co is 16 mol / cubic metre, how long does the reaction required to achieve the final concentration of 1 mol / cubic metre?
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Answer / kang chuen tat (malaysia - pen
Answer 52 - (a) dC / dt = -kC, dC / C = -k dt. Integrate both sides gives ln C - ln Co = ln (C / Co) = -kt, C = Co exp (-kt). (b) When C / Co = 0.5, t = [ ln (C / Co) ] / (-k) = [ ln (0.5) ] / (-0.139) = 4.9867 minutes. (c) Step 1 : 16 to 8 mol per cubic metre. Step 2 : 8 to 4. Step 3 : 4 to 2. Step 4 : 2 to 1. Each step takes t minutes. There are 4 steps needed from 16 to 1 mol / cubic metre or 4t = 19.9468 minutes. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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