Question

Two substances A $\left( {{t_{1/2}} = 5\min } \right)$ and B $\left( {{t_{1/2}} = 15\min } \right)$ follow first order kinetics are taken in such a way that initially $\left[ A \right] = 4\left[ B \right]$. Calculate the time after which the concentration of both the substances will be equal.
A) 15 min
B) 20 min
C) 24 min
D) 30 min

Answer 1

The substances A and B follow decay as first-order kinetics which means they both decay at the same rate with their respective half-life time. One can relate the relationship $\left[ A \right] = 4\left[ B \right]$ with the half-life time of the A and B and calculate the time where they both will have the same concentration.

Complete step by step answer:

1. First of all we will analyze the terms given in the question. The ratio between the two substances A and B has been given as $\left[ A \right] = 4\left[ B \right]$ which means we can use this relation while comparing the half-life time taken by it. Suppose we take the amount of substance B as ‘x’ we get,
   $\left[ A \right] = 4\left[ B \right]$
   If we take $B = x$ then,
   $\left[ A \right] = 4\left[ x \right]$
   We can write this as,
   $\left[ A \right] = 4x$
   Now let us take this value of A as ${\text{4x}}$ and the value of B as ${\text{x}}$.
2. Now as it is given in the question that the half-life of substance A is $\left( {{t_{1/2}} = 5\min } \right)$ and for substance B is $\left( {{t_{1/2}} = 15\min } \right)$. As the substance decays for the first time the amount ${\text{4x}}$ will become ${\text{2x}}$ after five minutes and after another decay, the amount will become from ${\text{2x}}$ to ${\text{x}}$ which is the second decay after ten minutes. After the third decay the amount from ${\text{x}}$ to $\dfrac{x}{2}$ after the fifteen minutes.
3. Now in the case of substance B which has a half-life of $\left( {{t_{1/2}} = 15\min } \right)$ which has amount as ${\text{x}}$. After the first decay which means after the fifteen minutes of time the initial amount of ${\text{x}}$ will become as $\dfrac{x}{2}$.
4. Now as we have seen in the above points that the final amount $\dfrac{x}{2}$ will become the same for substance A after three decays that are ${\text{15 minutes}}$ and for substance B after the first decay that is ${\text{15 minutes}}$. This means that both substances will be equal ${\text{15 minutes}}$.
   Therefore, substance A and B will become equal after ${\text{15 minutes}}$ which shows option A as the correct choice.

Note:
One can also find out the answer by using the formula ${C_t} = {C_0}{e^{ - Kt}}$ where ${C_t}$ the is concentration after time t and ${C_0}$ is the initial concentration. One can put the value of substance A and B by using this formula and put them in relation $\left[ A \right] = 4\left[ B \right]$. A half-life means the initial amount becomes half after the first decay and after each decay, the amount becomes half of the value which is calculated after the previous decay.