Question

A flask contains a mixture of compounds $A$ and $B$. Both compounds decompose by first order kinetics. The half-lives for $A$ and $B$ are $300s$ and $180s$, respectively. If the concentrations of $A$ and $B$ are equal initially, the time required for the concentration of $A$ to be four times that of $B$ (in $s$) is: (Use $\ln 2=0.693$)

• A. $180$
• B. $300$
• C. $120$
• D. $900$

Answer 1

The correct option is D

$900$

First-order Reaction:

1. A first-order reaction is a reaction in which the rate of reaction is proportional to the concentration of one reactant.
2. The half-life of a chemical reaction is the time taken for the initial concentration of reactants to become half of the original value.
3. It is given in that two compounds $A$and $B$ decompose by first-order kinetics
4. The half-life of $A$, $t_{1/2}=300s$ and half-life of $B$, $t_{1/2}=180s$.
5. When $t=t_{1/2}$ the concentration of reactant $\left[A\right]={\left[A\right]}_0{e}^{-k{t}_{1/2}}$.
6. Here, $A_0$is the initial concentration of reactant $A$
7. Replacing the given values of in the formula ${\left[\mathrm{A}\right]}_0{e}^{-k{t}_{1/2}}={\left[B\right]}_0{e}^{-k{t}_{1/2}}$.
8. ${\left[A\right]}_0{e}^{-\left(ln2/300\right)}=4{\left[B\right]}_0{e}^{-\left(\ln 2/180\right)}$
9. On solving the equation we get:$\frac{{\left[A\right]}_0{e}^{-\left(\ln 2/300\right)}}{{\left[B\right]}_0{e}^{-\left(\ln 2/180\right)}}=4$
10. Here it is given that the initial concentration of A and B are equal. so they cancel each other and the equation remains $\left[\frac{\ln 2}{180}-\frac{\ln 2}{300}\right]t=\ln 4$
11. Solving the above equation we get $\left[\frac{1}{180}-\frac{1}{300}\right]t=2$
12. This gives $t=\frac{2\times 180\times 300}{120}=900sec$.
13. So, at 900 sec the concentration of A is four times the concentration of B.
14. Hence, the correct option is D.