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 5.80×104 s1? Express your answer with the appropriate units.

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Part B What is the rate constant of a first-order reaction that takes 262 seconds for the reactant concentration to drop to half of its initial value? Express your answer with the appropriate units. -------------

Part C A certain first-order reaction has a rate constant of 7.90×103 s1. How long will it take for the reactant concentration to drop to 18 of its initial value?

Express your answer with the appropriate units.

Explanation / Answer

A)

if first order

t1/2 = 0.693/k

if k = 5.8*10^-4

then

t1/2 = (0.693)/(5.8*10^-4) = 1194.82758621 seconds

B)

find rate constant for

t1/2 = 262 s

t1/2 = 0.693/k

k = 0.693/262 = 0.00264559

K = 0.002645 or 2.65*10^-3

c)

k = 7.9*10^-3

to 1/8 of its value

1/2)^n = 1/8

n = 3

then

3 half lifes

t1/2 = (0.693/(7.9*10^-3) = 87.7215

total = 87.7215*3

t = 263.1645 seconds for 1/8

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