Half-life equation for first-order reactions: t1/2=0.693k where t1/2 is the half

Question

Half-life equation for first-order reactions: t1/2=0.693k where t1/2 is the half-life in seconds (s), and k is the rate constant in inverse seconds (s1). Part A What is the half-life of a first-order reaction with a rate constant of 6.60×104 s1? Part B What is the rate constant of a first-order reaction that takes 7.70 minutes for the reactant concentration to drop to half of its initial value? Part C A certain first-order reaction has a rate constant of 6.70×103 s1. How long will it take for the reactant concentration to drop to 1/8 of its initial value?

Explanation / Answer

Part A

K = 0.693/T1/2

T1/2 = 0.693/(6.6*10^(-4))   = 1050 sec

Part B

K = 0.693/7.7 = 0.09 min-1.

Part C

K = 6.7*10^(-3) s-1.

K = 2.303/tlog(a/a-x)

6.7*10^(-3) = 2.303/tlog(1/(1/8))

t = 310.42 sec

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