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?

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.

20 drops hope this

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