Question

A flask contains a mixture of compounds A and B. Both compounds decompose by first order kinetics. The half-lives are 50.0 minutes for A and 18.0 minutes for B. If the concentrations of A and B are equal initially, how long will it take for the concentration of A to be four times that of B?

Answer 1

Solution:-  

The integrated rate equation for systems following first-order kinetics is:  

N(t) = No × exp (-λ× t)  

where λ is the decay constant

which is related to the halflife, T,

λ = ln(2)/T  

so:  

N(t) = No×exp(-ln(2)×t/T)  

where No is the initial amount of N at time t = 0.  

In this case, we have for A and B:  

A(t) = Ao×exp(-ln(2)×t/(50 min))  

B(t) = Bo×exp(-ln(2)×t/(18 min))  

We want to know when 4×B(t) = A(t),

furthermore, we are told that Ao = Bo, so we want to know when:  

Ao×exp(-ln(2)*t/(50 min)) = 4×Ao×exp(-ln(2)×t/(18 min))  

Ao×exp(-ln(2)×t/(50 min)) = exp(ln(4))×Ao×exp(-ln(2)×t/(18 min))  

Ao×exp(-ln(2)×t/(50 min)) = Ao×exp(-ln(2)×t/(18 min) + ln(4))  

ln(2)×t/(50 min) = ln(2)×t/(18 min) - ln(4)  

ln(4)/ln(2) = t×(1/(18 min) - 1/(50 min))  

t = (ln(4)/ln(2))/(1/(18 min) - 1/(50 min))  

t = 56.25 min

Long will it take for the concentration of A to be four times that of B is 56.52 min

==================

@GauravSaxena01