The initial conditions are:

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

Half-life of A, $\left[\text{t}_{1/2}\right]_{A} = 10 \text{ min}$

Half-life of B, $\left[\text{t}_{1/2}\right]_{B} = 40 \text{ min}$

Both reactants follow first-order kinetics, and we want to find the time, $t$, when $[A]_t = [B]_t$.

Step-by-step Derivation:

Write the First-order Rate Equation:

For first-order reactions, the concentration $[X]_t$ at time $t$ is given by:

$ [X]_t = [X]_0 e^{-k_X t} $

where $k_X$ is the rate constant for substance $X$.

Express Half-life through Rate Constant:

For first-order reactions, the rate constant $k$ is related to the half-life $\text{t}_{1/2}$ by:

$ k = \frac{\ln 2}{\text{t}_{1/2}} $

Substituting the known half-lives:

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

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

Set Up the Equality from the Condition $[A]_t = [B]_t$:

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

By inserting the initial condition $[A]_0 = 8[B]_0$, it becomes:

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

Simplify and Solve for $t$:

Divide both sides by $[B]_0$:

$ 8 e^{-k_A t} = e^{-k_B t} $

Taking the natural logarithm of both sides:

$ \ln 8 = -k_A t + k_B t $

Replace $k_A$ and $k_B$ with their expressions:

$ \ln 8 = \left(\frac{\ln 2}{10} - \frac{\ln 2}{40}\right) t $

Simplify Further:

$ \ln 8 = \ln 2 \left(\frac{1}{10} - \frac{1}{40}\right) t $

Simplify $\left(\frac{1}{10} - \frac{1}{40}\right)$ to get:

$ \frac{1}{10} - \frac{1}{40} = \frac{4 - 1}{40} = \frac{3}{40} $

So,

$ \ln 8 = \ln 2 \times \frac{3}{40} \times t $

Final Expression for $t$:

$ t = \frac{\ln 8}{\ln 2 \times \frac{3}{40}} $

Calculate the value of $t$ using $\ln 8 = 3 \ln 2$:

$ t = \frac{3 \ln 2}{\ln 2 \times \frac{3}{40}} = \frac{40}{1} = 40 \text{ min} $

Therefore, after 40 minutes, the concentration of both reactants A and B will be the same.