Kerala KEAM 5 June 2024 Solved Paper

The integrated rate law for a first-order reaction is given by:$$\ln \left( \frac{[A]_0}{[A]} \right) = kt$$where:- $[A]_0$ is the initial concentration,- $[A]$ is the concentration at time $t$,- $k$ is the rate constant,- $t$ is the time.We are given: $$[A]_0 = 2 \, mol L^{-1}, \quad [A] = 0.125 \, mol L^{-1}, \quad t = 1 \, hr = 60 \, min.$$Substituting the values into the integrated rate law:$$\ln \left( \frac{2}{0.125} \right) = k \times 60$$Calculating the left-hand side:$$\ln \left( 16 \right) = k \times 60 \quad \Rightarrow \quad 2.7726 = k \times 60$$Solving for $k$:$$k = \frac{2.7726}{60} = 0.04621 \, min^{-1}$$Now, we use the formula for the half-life of a first-order reaction:$$t_{1/2} = \frac{\ln 2}{k}$$Substituting the value of $k$:$$t_{1/2} = \frac{0.693}{0.04621} \approx 15 \, min$$Thus, the half-life period of the reaction at 400K is $\boxed{15 \, min}$. Quick Tip: For a first-order reaction, use the integrated rate law to calculate the rate constant $k$ and then apply the half-life formula. The half-life of a first-order reaction is independent of the initial concentration.