Question

Chemical A is transformed into chemical B with the rate at which B is formed varying directly with the amount of A that is present at any instant in time. If there is 20 kg of chemical A present at the beginning of the process, and it is known that it takes 2 kg of A to produce 1 kg of B, then find the amount of chemical B that is formed as a function of time, if after 30 minutes, it is known that 4 kg of chemical B has been formed. If the reaction begins at 08:00, then at what time has 80% of chemical A been transformed into chemical B?

Answer 1

A B

Rate = k[A]

This is a first order reaction, rate depends on concentration one reactant. The rate constant for first order reaction can be calculated as follow:

k = 2.303/t log ([A_o] / [A_o - x]

[A_o] = 20kg and x = 4 kg

t = 30 min.

k = 2.303 /30 log(20/16)

k = 0.00743 min^-1

After 80% decomposition, 20 % A left

[A_o - x] = 20 % of 2kg

Using the above equation for k, calculate time t of 80% decomposition.

t = 2.303/k log ([A_o] / [A_o - x]

t = 2.303/0.00743 log (20/4)

t = 216.6 minutes