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Chapter 14: Problem 70

A flask contains a mixture of compounds \(A\) and \(B\). Both compounds decompose by first-order kinetics. The half-lives are 50.0 min for \(A\) and 18.0 min for \(B\). If the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are equal initially, how long will it take for the concentration of \(\mathrm{A}\) to be four times that of \(\mathrm{B} ?\)

Short Answer

Expert verified
It will take 56.25 minutes for the concentration of A to be four times that of B.

Step by step solution

01

Formulate the Problem

The problem asks for the time it takes for the concentration of A to be four times that of B. It needs to be solved as follows: \(C_{A} = 4C_{B}\)\nLet the initial concentrations of A and B be \(C_{0}\), then the concentrations at time \(t\) are given by: \n\(C_{A} = C_0 \times \frac{1}{2}^{(t/50)}\) for A and \(C_{B} = C_0 \times \frac{1}{2}^{(t/18)}\) for B.
02

Solve the Equality

In order to find the time when the concentration of A is four times that of B, we can set up the following equality and solve for \(t\). Set \(C_{A} = 4C_{B}\) and solve for \(t\). Plug in the specific expressions for \(C_{A}\) and \(C_{B}\) from step 1 into this equality, giving: \(C_0 \times \frac{1}{2}^{(t/50)} = 4 \times (C_0 \times \frac{1}{2}^{(t/18)})\). Since the initial concentration of A and B are the same, \(C_0\) cancels on both sides of the equation.
03

Solve for Time

The equation simplifies to \(\frac{1}{2}^{(t/50)} = 4 \times \frac{1}{2}^{(t/18)}\). Raise each side to the base 2, giving \(2^{t/50} = 2^{2 + t/18}\). Since the bases are the same, the exponents must be equal, giving \(t/50 = 2 + t/18\). Now solve for \(t\): Multiply through by 900 to clean up the fractions, providing 18t = 1800 + 50t. Subtract 18t from both sides: 1800 = 32t. Divide by 32 to isolate \(t\), giving \(t = 56.25\) minutes.

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