Half-life equation for first-order reactions: t 1/2=0.693/ k where t 1/2 is the

Question

Half-life equation for first-order reactions:
t1/2=0.693/k  
where t1/2 is the half-life in seconds (s), and k is the rate constant in inverse seconds (s1).

Part B What is the rate constant of a first-order reaction that takes 498 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 1.80×103 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

Half-life equation for first-order reactions:
t1/2=0.693/k  

Part B

t1/2 = 498 seconds

k = 0.693/t1/2

= 0.693/498

= 0.00139 S-1

Part C

k = 1.80×103 s1

k = 2.303/t1/2 log(a/a-x), a + initial value = 100, x = final value = 18

1.80×103 =1/t1/2 * 2.303log(100/100-18)

1.80×103 = 1/t1/2 * 2.303 * 0.0861

t1/2 = 110.16 seconds

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