Question

Thermal decomposition of a compound X follows first order kinetics. The initial concentration of X is 2 mol L\(^{-1}\). It decreased to 0.125 mol L\(^{-1}\) in one hour at 400K. What is the half-life period of the reaction at 400K? (log 2 = 0.3010)

• A. 15 min
• B. 20 min
• C. 30 min
• D. 25 min
• E. 10 min

Hint:

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.

Answer 1

The Correct Option is A

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 \( \boxed{15 \, {min}} \).