Cyclpropane is converted to its isomer propylene, when heated. The rate law is first order in cyclopropane and the rate constant is 6.0X10^-4/s at 500 degrees C. If the intial concentration of cyclopropane is 0.0226 mol/L, what is the concentration after 899s?

Answer 1

Isn't the first order equation
ln(Ao/A) = kt
Then ln(0.0226/A) = kt.
Substitute the k value and time of 899 seconds, and solve for A. Check my thinking.