Question

Two substances $A(t_{\frac{1}{2}}=5\text{min})$ and $B(t_{\frac{1}{2}}=15\text{min})$ are taken in such a way that initially $\left[A\right]=4\left[B\right]$. The time after which both the concentration will be equal is: (Assume that reaction is first order)

• A. $5min$
• B. $15min$
• C. $20min$
• D. Concentration can never be equal

Answer 1

The correct option is B $15min$

We know, for a first order reaction,
initial concentration and concentration at time $t$ is related as,
$C_t=C_0{e}^{-kt}$
According to question at time $t$
$C_{A_t}=C_{B_t}$
then,
$C_{A_0}{e}^{-k_{A}t}=C_{B_0}{e}^{-k_{B}t}$
$\frac{C_{A_0}}{C_{B_0}}=\frac{e^{-k_{B}t}}{e^{-k_{A}t}}\Rightarrow e^{(k_A-k_B)t}$
Given,
Initial concentration of $A$ is four times the initial concentration of $B$.
Half life of $A=5min$ and half life of $B=15min$
Also $k=\frac{ln2}{t_{\frac{1}{2}}}$ where $t_{\frac{1}{2}}$ is the half life of the reaction.

$4=e^{[\frac{ln2}{5}-\frac{ln2}{15}]\times t}$
$ln4=[\frac{ln2}{5}-\frac{ln2}{15}]t$
$ln(2{)}^2=[\frac{ln2}{5}-\frac{ln2}{15}]t$
$2ln2=[\frac{ln2}{5}-\frac{ln2}{15}]t$
$2=[\frac{1}{5}-\frac{1}{15}]t$
$2=\frac{2}{15}\times t$
$t=15minute$