Question

$A$ and $B$ are two different chemical species undergoing first order decompositions with half-life periods as $3$ and $4.5$ minutes respectively. If the initial concentrations of $A$ and $B$ are in the ratio $1:2$, The ratio $C_{t_{\left(A\right)}}/C_{t_{\left(B\right)}}$ after three half-lives of $A$ would be:

• A. $3:4$
• B. $1:1$
• C. $1:4$
• D. $4:3$

Answer 1

The correct option is C $1:4$

For the first order reaction, $C_t=C_o{\left(\frac{1}{2}\right)}^n$

where, $n=\frac{t}{t_{1/2}}$, $C_o=$ Initial concentration

After $3$ half- lives of $A$, ${C_t}_{\left(A\right)}={C_o}_{\left(A\right)}{\left(\frac{1}{2}\right)}^{\frac{3t_{1/2}}{t_{1/2}}}=\frac{{C_o}_{\left(A\right)}}{8}$

After $3$ half -lives of $A$, for $B$, $n=\frac{3\times 3}{4.5}=2$

$\therefore {C_t}_{\left(B\right)}={C_o}_{\left(B\right)}{\left(\frac{1}{2}\right)}^2=\frac{{C_o}_{\left(B\right)}}{4}$

$\therefore \frac{{C_t}_{\left(A\right)}}{{C_t}_{\left(B\right)}}=\left(\frac{{C_o}_{\left(A\right)}}{{C_o}_{\left(B\right)}}\right)\times \frac{4}{8}=\frac{1}{2}\times \frac{4}{8}=1:4$