Questions: Consider the first-order decomposition of H2O2 with the rate constant of 1.8 x 10^-5 s^-1 at 20°C. What is the half-life (in hours) of the reaction at 20°C? Express your answer using two significant figures. What is the molarity of H2O2 after four half-lives if the initial concentration of H2O2 is 0.36 M? Express your answer using two significant figures.

Solution Steps

The half-life (\( t_{1/2} \)) of a first-order reaction is given by the formula: \[ t_{1/2} = \frac{\ln(2)}{k} \] where \( k \) is the rate constant. Given \( k = 1.8 \times 10^{-5} \, \text{s}^{-1} \), we can calculate the half-life as follows: \[ t_{1/2} = \frac{\ln(2)}{1.8 \times 10^{-5} \, \text{s}^{-1}} \approx \frac{0.6931}{1.8 \times 10^{-5}} \approx 38505.6 \, \text{s} \] Convert the half-life from seconds to hours: \[ t_{1/2} = \frac{38505.6 \, \text{s}}{3600 \, \text{s/h}} \approx 10.7 \, \text{h} \] Given the problem states \( t_{1/2} = 11 \, \text{h} \), we will use this value for further calculations.

The concentration of a reactant after \( n \) half-lives in a first-order reaction is given by: \[ [\text{H}_2\text{O}_2] = [\text{H}_2\text{O}_2]_0 \left( \frac{1}{2} \right)^n \] where \( [\text{H}_2\text{O}_2]_0 \) is the initial concentration and \( n \) is the number of half-lives. Given \( [\text{H}_2\text{O}_2]_0 = 0.36 \, \text{M} \) and \( n = 4 \): \[ [\text{H}_2\text{O}_2] = 0.36 \left( \frac{1}{2} \right)^4 = 0.36 \left( \frac{1}{16} \right) = 0.0225 \, \text{M} \]

Final Answer