StemCET Logo

Chemistry Question 58 – JEE-MAIN 2025

← Back
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 following first order kinetics, after how much time will the concentration of both the reactants be same ?

For first-order reactions, the concentration of a reactant decreases exponentially with time, and its half-life is constant.

Step 1: Calculate rate constants for A and B✦ Active

For a first-order reaction, the rate constant k is related to the half-life t1/2 by the formula k=0.693t1/2. Using the given half-lives:

kA=0.69310 minandkB=0.69340 min
Step 2: Set up and equate concentration expressions○ Expand

The concentration of a reactant at time t for a first-order reaction is given by [X]t=[X]0e−kt. We are given [A]0=8[B]0 and need to find the time t when [A]t=[B]t.

[A]0e−kAt=[B]0e−kBt 8[B]0e−kAt=[B]0e−kBt 8e−kAt=e−kBt
Step 3: Solve for time t○ Expand

Rearrange the equation and take the natural logarithm to solve for t.

8=e−kBte−kAt=e(kA−kB)t ln⁡(8)=(kA−kB)t t=ln⁡(8)kA−kB t=3ln⁡(2)0.69310−0.69340=3ln⁡(2)0.693(110−140) t=3ln⁡(2)0.693(4−140)=3ln⁡(2)0.693(340)

Since 0.693≈ln⁡(2), we can simplify:

t=3ln⁡(2)ln⁡(2)(340)=3340=3×403=40 min
💡 Teacher's Secret Hint

Remember that ln⁡(8) can be written as 3ln⁡(2) to simplify calculations with 0.693.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.