Half-life chem problem. please help.?

Question

C4H8 → 2C2H4

The half-life of this first-order reaction is 22.7 s. If the initial concentration of C4H8 is 1.42 M, what will be the rate of reaction at t = 60 s?

Answer 1

To solve this problem, it is mostly just memorization of rate laws. First we can find out the rate constant, since they give you the half life. Using the equation:

k= rate constant t1/2= half life

ln2 / k = t1/2

22.7s = ln2/k

k= .03054s^-1

Next we can find the concentration of the reactant at the time of 60s, using the integrated first-order rate law.

[A] = final concentration of C4H8 [A]0 = initial concentration of C4H8

t = time elapsed k= rate constant

ln[A] = -kt + ln[A]0

ln[A] = (.03054s^-1)*(60s) + ln(1.42M)

ln[A] = -1.481

[A] = .227M

Now we have the concentration of C4H8 after 60s. Now we use the rate law to find the rate at 60s.

k= rate constant

Rate = [C4H8]*k (don't for get the rate law only deals with reactants)

Rate = (.227M)*(.0305s^-1)

Rate = 6.94*10^-3 M/s  

Answer 2

t 1/2 = ln 2 / k

22.7 = 0.693 / k

k = 0.0305

ln [C4H8]f = - 0.0305 x 60 + ln 1.42 = - 1.83 +  0.351 =  - 1.48

e^- 1.48 = 0.229 M = [C4H8]f