Chemical A is transformed into chemical B The rate at which

Chemical A is transformed into chemical B. The rate at which B is formed varies directly with the amount of A present at any instant. if 10lb of A is present initially and 3lb of A is transformed into B in 1hr, find the a)amount of A transformed after 2,3and 4 hours, b)time in which 75% of chemical A would be transformed to chemical B.

Solution

The rate at which B is formed varies directly with the amount of A present at any instant. This gives the differential equation dB/dt = (10 – A) where is a constant of proportionality

The general solution is B = (10 – A)t + k where k is an arbitrary constant

At t = 0, B = 0, gives k = 0. So the solution becomes B = (10 – A)t

At t = 1, A = B = 3, gives 3 = 7. So, = 3/7

The solution becomes, B = 3(10 – A)t/7 or, A = 3(10 – A)t/7 because at any point of time t, A = B

Simplifying the equation, 7A = 3(10 – A)t

a) when t = 2, we get 7A = 6(10 – A)

or, 13A = 60

or, A = 60/13

when t = 3, we get 7A = 9(10 – A)

or, 16A = 90

or, 8A = 45

or, A = 45/8

when t = 4, we get 7A = 12(10 – A)

or, 19A = 120

or, A = 120/19

Therefore, amount of A transformed after 2,3 and 4 hours is give as 60/13 lb, 45/8 lb and 120/19 lb respectively. (Answer)

b) To find time t at which 75% of A will be transformed into B, i.e. substitute A = 7.5 in the equation 7A = 3(10 – A)t

7 x 7.5 = 3(10 – 7.5)t

Or, 52.5 = 7.5t

Or, t = 7

Therefore, after 7 hours, 75% of chemical A would be transformed to chemical B. (Answer)