Question

A and B are two different chemical species undergoing 1st order decomposition with half lives equal to $5$ sec. and $7.5$ sec. respectively. If the initial concentration of A and B are in the ratio $3:2$, calculate $\frac{C_A}{C_B}$, after three half lives of A. Report your answer after multiplying it with $100$.

Answer 1

$t_{1/2}\left(A\right)=5sec=\frac{0.693}{k_A}{k}_A=0.1386{sec}^{-1}{t}_{1/2}\left(B\right)=1.5sec=\frac{0.693}{k_B}{k}_B=0.0924{sec}^{-1}$

Let initial concentration of A be 150.

Then initial concentration of B is 100.

Three half lives of $A=5\times 3=15sec.C_A=\frac{150}{2^3}=18.75mol/L$

$15sec=2$ half lives of B

$C_B=\frac{100}{2^2}=25mol/L\frac{C_A}{C_B}=\frac{18.75}{25}=0.75$

Multiplying it by 100 we get

$0.75\times 100=75$