How many seconds will it take for the [C5H6] to be 0.110 M? The k for the reaction is 0.267 s–1.

Question

Cyclopentadiene (C₅H₆) reacts with itself to form dicyclopentadiene (C₁₀H₁₂). A 0.880 M solution of C₅H₆ is monitored as a function of time as the reaction proceeds. A graph of ln[C₅H₆] versus time yields a straight line. 

How many seconds will it take for the [C5H6] to be 0.110 M? The k for the reaction is 0.267 s–1.

Answer 1

Great question! So to start off, they tell us that the graph of ln[C₅H₆] vs time yields a straight line. This must mean that the reaction is a first order reaction as the the ln is the identifying component of that equation. The formula for a first order reaction is below:

ln[A]_t = -kt + ln[A]₀

where [A]_t is the concentration at time t and [A]₀ is the initial concentration. Now its a manner of plugging in the values we know:

ln[C₅H₆]₀ = 0.880 M

ln[C₅H₆]_t = 0.110 M

k = 0.267 s^-1

ln[0.110] = -(0.267)t + ln[0.880]

ln[0.110] - ln[0.880] = -(0.267)t

-2.079 = -(0.267)t

t = (-2.079)/(-0.267)

t = 7.788 s

The time it would take for the concentration to be 0.110 M is 7.788 s. Please let me know if that made sense and if you have any other questions!