Butadiene dimerises by a second order path with k = 5.76 x 10^-2 L/mol/min. Starting at 0.200 M, what concentration remains after 10.0 min?

1. The second order integrated law is 1/[A] = kt + 1/[A]0. 2. The term kt is 5.76 x 10^-2 L/mol/min times 10.0 min, which is 0.576 L/mol. 3. The term 1/[A]0 is 1 divided by 0.200 mol/L, which is 5.00 L/mol. 4. Adding gives 1/[A] = 5.576 L/mol. 5. Taking the reciprocal gives [A] = 0.179 mol/L. 6. Choosing 0.100 mol/L assumes one half-life has passed, but the half-life here is 1/(k[A]0), which is about 87 min. 7. Choosing 0.200 mol/L leaves the concentration untouched and ignores the elapsed time entirely. _Source: OpenStax Chemistry (CC BY 4.0), Ch 12 "Kinetics", section 12.4 Integrated Rate Laws_