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 1.80×104 s1?Express your answer with the appropriate units.

Part B What is the rate constant of a first-order reaction that takes 369 secondsfor 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 7.50×103 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

t1/2=0.693/k

a) t1/2=0.693/(1.8*10^-4) =3850s

b) k=0.693/t1/2

k =0.693/369 =0.001878=1.878*10^-3s^-1

c) k=7.50*10^3

one t1/2 for concentration to get 1/2 of initial value

one more for concentration to get 1/2 of above value (i.e 1/2 of 1/2 of initial value=1/4 of initial value)

one more for concentration to get 1/8 of initialvalue

=> total time taken =3*t1/2

where, t1/2 =0.693/k =0.693/7.50*10^3 =92.4s

=> total time =3*92.4 =277.2s

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