Answer to Question #191000 in Differential Equations for Manahil Faisal

Question

In a chemical manufacturing plant, a certain type of chemical B is produced

through reactions in chemical A. Through observation the officers on duty have

noted that the rate of conversion from chemical A to B is proportion to the

amount of chemical A present at any time, further investigation has revealed

that 10% amount has been converted in first five minutes.

a) The plant manager wishes to know the percentage of chemical A that will

be converted in 20 minutes.

b) The plant manager wishes to know the time it will take to convert 60

minutes of chemical A.

Answer 1

For this

We first calculate t_1/2 that is half conversion time

${N(t)\over N_o} = ({1\over 2})^{t\over t_{1\over 2}}$ .......(1)

$N(t)=$ amount at time t (let it be 100)

$N_o=$ initial amount

By putting values

Given t = 5 min

Amount converted = 10%

So

$N_o=$ 90

(Note we use ln to solve)

We got

t_1/2=34.5

a) at t=20 and t_1/2=34.5

Putting values in 1 we got

${N(20)\over N_o}=({1\over 2})^{20\over 34.5}$

And solving

We got

N(20) = 67.3 %

Converted = 100-67.3

=32.7%

b)

So N(t)=100-60

=40%

N_o= 100

t_1/2=34.5

Putting values in (1)

And we got

${40\over100 }=({1\over 2})^{t\over {34.5}}$

taking ln both side , we got

t = 45.5 minutes