The decomposition of XY is second order in XY and has a rate constant of 7.02 * 10^-3 M^-1 s^-1 at a certain temperature. b. How long will it take for the concentration of XY to decrease to 12.5% of its initial concentration when the initial concentration is 0.100 M? When the initial concentration is 0.200 M? c. If the initial concentration of XY is 0.150 M, how long will it take for the concentration to decrease to 0.062 M? d. If the initial concentration of XY is 0.050 M, what is the concentration of XY after 5.0 * 10^1 s? After 5.50 * 10^2 s?

Verified step by step guidance

1. Identify that the reaction is second order, which means the rate law is given by \( \text{Rate} = k[XY]^2 \).
2. Use the integrated rate law for a second-order reaction: \( \frac{1}{[XY]} = kt + \frac{1}{[XY]_0} \), where \([XY]_0\) is the initial concentration and \([XY]\) is the concentration at time \(t\).
3. For part b, substitute \([XY]_0 = 0.100 \text{ M}\) and \([XY] = 0.0125 \text{ M}\) into the integrated rate law to solve for \(t\). Repeat the process for \([XY]_0 = 0.200 \text{ M}\).
4. For part c, substitute \([XY]_0 = 0.150 \text{ M}\) and \([XY] = 0.062 \text{ M}\) into the integrated rate law to solve for \(t\).
5. For part d, use the integrated rate law with \([XY]_0 = 0.050 \text{ M}\) to find \([XY]\) after \(t = 50 \text{ s}\) and \(t = 550 \text{ s}\).