A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. In what time the reaction will be 80% complete?

1) For the first order reaction:
k = (2.303 / t) log (a / a – x)
When x = (40 / 100) a = 0.4 a
t = 50 minutes (given)
Therefore, k = (2.303 / 50) log (a / a – 0.4 a)
k = (2.303 / 50) log (1 / 0.6)
= 0.010216 min-1
Hence the value of the rate constant is 0.010216 min-1

2) t= ?, when x = 0.8 a
From above, k = 0.010216 min-1
Therefore, t = (2.303 / 0.010216) log (a / a – 0.8 a)
= (2.303 / 0.010216) log (1 / 0.2) = 157.58 min
The time at which the reaction will be 80% complete is 157.58 min.

Category: Chemical Kinetics

More Questions

Copyright © All rights reserved. TheBigger.com | Privacy Policy | Contact Us | Copyright Policy | Useful Resources