Question

He rate law for the reaction 3 a c is rate = 4.36 ✕ 10-2 l mol-1 hr-1[a]2 what is the half-life for the reaction if the initial concentration of a is 0.250 m?

Answer 1

The half-life for the reaction can be determined using the second-order half-life equation: t1/2 = 1 / (k * [A]0), where t1/2 is the half-life, k is the rate constant, and [A]0 is the initial concentration of A. Substituting the given values into the equation, we find that the half-life is 144.93 hours.

Step-by-step explanation:

The half-life for the reaction can be determined using the second-order half-life equation:

t1/2 = 1 / (k * [A]0)

where t1/2 is the half-life, k is the rate constant, and [A]0 is the initial concentration of A.

Substituting the given values into the equation, we have:

t1/2 = 1 / (4.36 × 10-2 L mol-1 hr-1 × (0.250 M)2)

t1/2 = 144.93 hours

Answer 2

d[C]dt=.0436[A]2d[C]dt=.0436[A]2

d[A]dt=−3d[C]dtd[A]dt=−3d[C]dt

d[A]dt=−.1308[A]2d[A]dt=−.1308[A]2

−1[A]2d[A]=.1308dt−1[A]2d[A]=.1308dt

1[A]=.1308t+C1[A]=.1308t+C

[A]=1.1308t+C[A]=1.1308t+C

.250=1C.250=1C

C=4C=4

.125=1.1308t+4.125=1.1308t+4

t=(1.125−4)∗1.1308=30.58t=(1.125−4)∗1.1308=30.58

Assuming I did my math right, looks fine to me, it should be about 30 hours and 30 minutes.