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

Answer: (B)

Given data,

Initial concentration of $A$ is four times the initial concentration of $B$.

Half life of $A=5min$ and half life of $B=15min$

$\left[A\right]=4\left[B\right]$.

Also $k=\frac{ln2}{t_{\frac{1}{2}}}$ where $t_{\frac{1}{2}}$ is the half life of the reaction.

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,

$C_{A_0}{e}^{-k_{A}t}=C_{B_0}{e}^{-k_{B}t}$

$\frac{C_{A_0}}{C_{B_0}}=\frac{e^{-k_{B}t}}{e^{-k_{A}t}}\Rightarrow e^{(k_A-k_B)t}$

$4=e^{[\frac{ln2}{5}-\frac{ln2}{15}]\times t}$

$ln4=[\frac{ln2}{5}-\frac{ln2}{15}]t$

$ln(2{)}^2=[\frac{ln2}{5}-\frac{ln2}{15}]t$

$2ln2=[\frac{ln2}{5}-\frac{ln2}{15}]t$

$2=[\frac{1}{5}-\frac{1}{15}]t$

$2=\frac{2}{15}\times t$

$t=15minute$

Conclusion: The time after which both the concentration will be equal is 15 minutes. Option (B) is the correct answer.