COMEDK 2025 Morning Solved Paper

Concept:For a first-order reaction, the half-life is given by $t_{1/2} = \frac{\ln 2}{k}$, where $k$ is the rate constant. We first find $k$ from the given concentration change using the integrated rate law.Explanation:Step 1: Find the concentration of X after 200 minutes.The reaction is $\mathrm{X} \rightarrow 2\mathrm{Y}$.Given: initial [X] = $1.0\ \mathrm{mol/L}$, and [Y] formed = $0.4\ \mathrm{mol/L}$.Stoichiometry: 1 mol X produces 2 mol Y. So, moles of X consumed = $\frac{0.4}{2} = 0.2\ \mathrm{mol/L}$.Thus, [X] at time $t$ = $1.0 - 0.2 = 0.8\ \mathrm{mol/L}$.Step 2: Apply the first-order integrated rate law.For a first-order reaction: $\ln\frac{[\mathrm{X}]_t}{[\mathrm{X}]_0} = -k t$.Here, $[\mathrm{X}]_0 = 1.0$, $[\mathrm{X}]_t = 0.8$, $t = 200\ \mathrm{min}$.So, $\ln\left(\frac{0.8}{1.0}\right) = -k(200)$.Thus, $k = -\frac{1}{200} \ln(0.8) = \frac{1}{200} \ln\left(\frac{1}{0.8}\right)$.Calculate: $\ln(1.25) \approx 0.22314$, so $k \approx \frac{0.22314}{200} = 0.0011157\ \mathrm{min}^{-1}$.Step 3: Compute the half-life.$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693147}{0.0011157} \approx 620.96\ \mathrm{min}$.Answer:Option C ($620.96\ \mathrm{min}$).