Answer to Question #285715 in Chemical Engineering for Falak

Question

Liquid A decomposes by first-order kinetics, and in a batch reactor 50% of A is converted in a 5-minute run. How much longer would it take to reach 75% conversion?

Answer 1

This question may be answered by two different mechanisms. The first is by simple inspection. In each 5-minute period the amount of A is reduced by 1/2. That 5 minute period is termed a “half-life”. If 75% is converted, only 25% remains, which is 1/2 of 50%. Therefore, another 5-minutes are needed for A to decay from 50% down to 25%.

Or, you can do the math. Considering the amount A to be a function of time, you have

A(t) = A(0)*exp(-xt) We can find x using the natural log of both sides. From the original data, with t= 5 minutes: Ln[A(5)] = Ln(0.50) = -0.693

-0.693 = Ln[exp(-xt)] = -x*5 Now divide both sides by -5 and you get x = 0.138

Use that value of x to calculate the time for A to fall to 25%

0.25 = exp(-0.138*t), Taking the natural log of both sides gives Ln(0.25) = -0.138*t

Ln (0.25) = -1.38 = -0.138* t, so for A(t) to fall to 25% is two half lives, or 10 minutes.

Note: First order chemical reactions and radioactive decay are described by the same type of equation