In a reaction A + B C, initial concentrations of A and B are related as [A]0 = 8[B]0. The half lives of A and B are 10 min and 40 min, respectively. If they start to disappear at the same time, both…

Given: [A]0 = 8[B]0, t1/2(A) = 10 min, t1/2(B) = 40 min. Both reactants follow first order kinetics. Find: The time t when [A] = [B]. For first order decay, [A] = [A]0 e^-kA t and [B] = [B]0 e^-kB t At the required time, concentrations become equal: [A]0 e^-kA t = [B]0 e^-kB t Using [A]0 = 8[B]0, 8[B]0 e^-kA t = [B]0 e^-kB t So, 8 = e^(kA-kB)t Taking logarithm, 8 = (kA-kB)t Hence, t = 8kA-kB For a first order reaction, k = 2t1/2 Therefore, kA = 210 and kB = 240 Substituting, t = 8 210 - 240 Now simplify: 8 = 2^3 = 3 2 So, t = 3 2 2 ( 110- 140) = 3 4-140 = 3 340 = 40 Therefore, the concentration of both reactants becomes same after 40 min. The correct option is D.

In a reaction A + B C, initial concentrations of A and B are rela… — Solution | JEE Chemistry

MCQMediumJEE 2025Integrated Rate Laws

In a reaction $A + B \rightarrow C$ , initial concentrations of A and B are related as $[A]_0 = 8[B]_0$ . The half lives of A and B are $10 \, \text{min}$ and $40 \, \text{min}$ , respectively. If they start to disappear at the same time, both following first order kinetics, after how much time will the concentration of both the reactants be same?

A
$60 \, \text{min}$
B
$80 \, \text{min}$
C
$20 \, \text{min}$
D
$40 \, \text{min}$

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: $[A]_0 = 8[B]_0$ , $t_{1/2(A)} = 10 \, \text{min}$ , $t_{1/2(B)} = 40 \, \text{min}$ . Both reactants follow first order kinetics.

Find: The time $t$ when $[A] = [B]$ .

For first order decay,

[A] = [A]_0 e^{-k_A t}

and

[B] = [B]_0 e^{-k_B t}

At the required time, concentrations become equal:

[A]_0 e^{-k_A t} = [B]_0 e^{-k_B t}

Using $[A]_0 = 8[B]_0$ ,

8[B]_0 e^{-k_A t} = [B]_0 e^{-k_B t}

So,

8 = e^{(k_A-k_B)t}

Taking logarithm,

\ln 8 = (k_A-k_B)t

Hence,

t = \frac{\ln 8}{k_A-k_B}

For a first order reaction,

k = \frac{\ln 2}{t_{1/2}}

Therefore,

k_A = \frac{\ln 2}{10}

and

k_B = \frac{\ln 2}{40}

Substituting,

t = \frac{\ln 8}{\frac{\ln 2}{10} - \frac{\ln 2}{40}}

Now simplify:

\ln 8 = \ln 2^3 = 3\ln 2

So,

t = \frac{3\ln 2}{\ln 2\left(\frac{1}{10}-\frac{1}{40}\right)}

= \frac{3}{\frac{4-1}{40}}

= \frac{3}{\frac{3}{40}} = 40

Therefore, the concentration of both reactants becomes same after $40 \, \text{min}$ . The correct option is D.

Common mistakes

Using the same rate constant for A and B is incorrect because their half-lives are different. For first order kinetics, different half-lives imply different $k$ values. First calculate $k_A$ and $k_B$ separately using $k = \frac{\ln 2}{t_{1/2}}$ .
Equating half-lives instead of concentrations is wrong. The question asks for the time when $[A] = [B]$ , so you must write the concentration-time expressions and then set them equal.
Ignoring the initial relation $[A]_0 = 8[B]_0$ leads to an incorrect equation. The factor $8$ must be included when equating the concentration expressions; otherwise the time obtained will be wrong.

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Answer

Step-by-step solution

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