The decomposition of XY is second order in XY and has a rate constant of 7.02 x 10^-3 M^-1 s^-1 at a certain temperature. How long will it take for the concentration of XY to decrease to 12.5% of its initial concentration when the initial concentration is 0.80 M?

• A. 128 seconds
• B. 256 seconds
• C. 512 seconds
• D. 1024 seconds

Answer: (C)

Verified step by step guidance

1. Identify that the reaction is second order with respect to XY, which means the rate law is given by: $d\mathrm{[XY]}/dt=-k\cdot {\mathrm{[XY]}}^2$.
2. Use the integrated rate law for a second-order reaction: $(1/\mathrm{[XY]})=(1/\mathrm{[XY]})+k\cdot t$, where $\mathrm{[XY]}$ is the concentration at time $t$.
3. Determine the final concentration of XY, which is 12.5% of the initial concentration. Calculate this as: $\mathrm{[XY]}=0.125\cdot 0.80=0.10M$.
4. Substitute the known values into the integrated rate law: $(1/0.10)=(1/0.80)+7.02\cdot {10}^{-3}\cdot t$.
5. Solve for $t$ to find the time it takes for the concentration to decrease to 12.5% of its initial value.