Two substances A and B are present such that [A0] = 4[B0] and half-life of A is 6 minute and that of B is 15 minute. If they start decaying at the same time following first order kinetics, how much time (min) later will the concentration of both of them would be same?Correct answer is '15'. Can you explain this answer?

Question

Answer

Understanding the Problem

We are given two substances, A and B, with different half-lives. The initial concentrations of these substances are related by [A0] = 4[B0]. We need to determine the time at which the concentrations of A and B will become equal.

First Order Kinetics

The decay of a substance following first order kinetics can be described by the equation:

[A] = [A0] * e^(-kt)

where [A] is the concentration of the substance at a given time, [A0] is the initial concentration, k is the rate constant, and t is the time elapsed.

Half-Life

The half-life of a substance is the time it takes for half of the initial concentration to decay. For a first-order reaction, the half-life can be calculated using the equation:

t(1/2) = ln(2)/k

where t(1/2) is the half-life and ln represents the natural logarithm.

Relationship between Half-Life and Rate Constant

From the equation for the half-life, we can rearrange it to solve for the rate constant:

k = ln(2)/t(1/2)

This relationship tells us that substances with shorter half-lives have larger rate constants, indicating that they decay faster.

Initial Concentrations

We are given that [A0] = 4[B0]. Let's assume that [A0] = 4x and [B0] = x.

Therefore, [A0]/[B0] = 4x/x = 4.

Calculating the Rate Constants

Using the half-life equation, we can calculate the rate constants for substances A and B:

kA = ln(2)/tA(1/2) = ln(2)/6
kB = ln(2)/tB(1/2) = ln(2)/15

Calculating the Concentrations at a Given Time

Let's consider a time t when the concentrations of A and B become equal. At this time, [A] = [B].

[A] = [A0] * e^(-kA*t)
[B] = [B0] * e^(-kB*t)

Since [A] = [B], we can set up the following equation:

[A0] * e^(-kA*t) = [B0] * e^(-kB*t)

Substituting the initial concentrations, we have:

4x * e^(-kA*t) = x * e^(-kB*t)

Cancelling out the x terms, we obtain:

4 * e^(-kA*t) = e^(-kB*t)

Taking the natural logarithm of both sides, we have:

ln(4) - kA*t = -kB*t

Rearranging the equation, we get:

-kA*t + kB*t = ln(4)

Factoring out t, we have:

t * (kB - kA) = ln(4)

Simplifying further, we obtain:

t = ln(4) / (kB - kA)

Substituting the values of kA and kB, we have:

t = ln(4) / (ln(2)/15 - ln(2)/6)

Simplifying

Answer

I think answer should be 20

Answer

Answer

30

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