Chemistry question on rate laws

Question

For the second order reaction: 2C4H6(g) → C8H12(g) If 35.0% of 0.100 mol.dm–3 initial concentration has reacted after 2.00 hours, what is the half–life for the reaction?

Answer 1

2C₄H₆₍g) ==> C₈H₁₂(g)

Initial [C₄H₆] = 0.100 M

[C₄H₆] after 2.00 hours = 0.65 x 0.100 = 0.065 M

For a second order reaction, t_1/2 = 1/k[A]_o

We need to find the value of k, the rate constant.

1/[A] = 1/[A]_o + kt

1/0.065 = 1/0.100 + k(2 hrs) note: time is in hours

15.38 = 10 + 2k

5.38 = 2k

k = 2.69 M^-1hr^-1

t_1/2 = 1/k[A]_o

t_1/2 = 1/2.69 (0.100)

t_1/2 = 0.0372 hrs = 2.23 min = 134 sec