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 (s−1).

a) What is the half-life of a first-order reaction with a rate constant of 4.80×10⁻⁴  s−1?

b) What is the rate constant of a first-order reaction that takes 188 seconds for the reactant concentration to drop to half of its initial value?

Express your answer with the appropriate units.

c)A certain first-order reaction has a rate constant of 7.90×10⁻³ s−1. How long will it take for the reactant concentration to drop to 18 of its initial value?

Express your answer with the appropriate units.

Answer 1

Dropping to 18 of it intial value means at that time the concentration is 72

Half life, t1/2 = 0.693/k l = 4.8 times 10^-4 t1/2 = 0.693/4.8 times 10^-4 = 1433.755 t1/2 = 188s k = 0.693/t1/2 = 0.693/188 = 3.68 times 10\^-3 s^-1 k = 7.9 times 106-3 s^-1 For transfer reaction k = a0 = 100 at = 100-18 = 82 k = 1/t ln 100/82 t = 1/k ln 100/82 = 1000/7.9 times ln 100/82 = 25.125