The first-order reaction, 2 N2O(g) → 2 N2(g) + O2(g), has a rate constant equal to 0.76 s-1at 1000 K.

Question

How long will it take for the concentration of N2O to decrease to 42% of its initial concentration?

Answer 1

For a first order reaction, the half life is t_1/2 = 0.693/k

Therefore, t_1/2 = 0.693/0.76s^-1 = 0.9118 s

Now that we know the half life, a simple formula to use is FR = 0.5ⁿ where FR = fraction remaining and n = number of half lives that have elapsed. In the current problem, we want to find how many half lives it takes to reach 42% of the original amount or when the fraction remaining = 0.42

0.42 = 0.5ⁿ

log 0.42 = n log 0.5

-0.377 = -0.301 n

n = 1.25 half lives and since each half life is 0.9118 seconds, we can find the time...

time = 1.25 half lives x 0.9118 second/half life = 1.1 seconds