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 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?

<p>The initial conditions are:</p> <p><p>$[A]_0 = 8[B]_0$</p></p> <p><p>Half-life of A, $\left[\text{t}_{1/2}\right]_{A} = 10 \text{ min}$</p></p> <p><p>Half-life of B, $\left[\text{t}_{1/2}\right]_{B} = 40 \text{ min}$</p></p> <p>Both reactants follow first-order kinetics, and we want to find the time, $t$, when $[A]_t = [B]_t$.</p> <h3>Step-by-step Derivation:</h3> <p><p><strong>Write the First-order Rate Equation:</strong></p> <p>For first-order reactions, the concentration $[X]_t$ at time $t$ is given by:</p> <p>$ [X]_t = [X]_0 e^{-k_X t} $</p> <p>where $k_X$ is the rate constant for substance $X$.</p></p> <p><p><strong>Express Half-life through Rate Constant:</strong></p> <p>For first-order reactions, the rate constant $k$ is related to the half-life $\text{t}_{1/2}$ by:</p> <p>$ k = \frac{\ln 2}{\text{t}_{1/2}} $</p> <p>Substituting the known half-lives:</p></p> <p><p>$k_A = \frac{\ln 2}{10}$</p></p> <p><p>$k_B = \frac{\ln 2}{40}$</p></p> <p><p><strong>Set Up the Equality from the Condition $[A]_t = [B]_t$:</strong></p> <p>$ [A]_0 e^{-k_A t} = [B]_0 e^{-k_B t} $</p> <p>By inserting the initial condition $[A]_0 = 8[B]_0$, it becomes:</p> <p>$ 8[B]_0 e^{-k_A t} = [B]_0 e^{-k_B t} $</p></p> <p><p><strong>Simplify and Solve for $t$:</strong></p> <p>Divide both sides by $[B]_0$:</p> <p>$ 8 e^{-k_A t} = e^{-k_B t} $</p> <p>Taking the natural logarithm of both sides:</p> <p>$ \ln 8 = -k_A t + k_B t $</p> <p>Replace $k_A$ and $k_B$ with their expressions:</p> <p>$ \ln 8 = \left(\frac{\ln 2}{10} - \frac{\ln 2}{40}\right) t $</p></p> <p><p><strong>Simplify Further:</strong></p> <p>$ \ln 8 = \ln 2 \left(\frac{1}{10} - \frac{1}{40}\right) t $</p> <p>Simplify $\left(\frac{1}{10} - \frac{1}{40}\right)$ to get:</p> <p>$ \frac{1}{10} - \frac{1}{40} = \frac{4 - 1}{40} = \frac{3}{40} $</p> <p>So,</p> <p>$ \ln 8 = \ln 2 \times \frac{3}{40} \times t $</p></p> <p><p><strong>Final Expression for $t$:</strong></p> <p>$ t = \frac{\ln 8}{\ln 2 \times \frac{3}{40}} $</p> <p>Calculate the value of $t$ using $\ln 8 = 3 \ln 2$:</p> <p>$ t = \frac{3 \ln 2}{\ln 2 \times \frac{3}{40}} = \frac{40}{1} = 40 \text{ min} $</p></p> <p>Therefore, after 40 minutes, the concentration of both reactants A and B will be the same.</p>