Kinetics order chemistry

Question

A first order reaction has a half-life of 26.41 seconds. How long does it take for the concentration of the reactant in the reaction to fall to 63.25% of its initial value? How long does it take for the concentration of the reactant in the reaction to fall to 13.75% of its initial value?

Answer 1

for a first order reaction A → products we have d[A]/dt = –k[A] leading to

[A]/[A]₀ = e^–kt where t = time (in s) and k = the rate constant (in s^–1)

The value of t when [A]/[A]₀ = ½ is called the half life of the reaction and is represented as t_½

The relationship between half life (t_½) and rate constant (k) can be found by writing the expression

½ = e^–k(t½) and solving for t_½

Taking the natural logarithm of both sides gives ln(½) = –ln2 = –kt_½ which means that

t_½ = ln2/k or k = ln2/t_½

In the problem we're given t_½ = 26.41 s, so k = ln2/21.41 = 3.237 × 10^–2 s^–1

Question 1

How long does it take for the concentration of the reactant in the reaction to fall to 63.25% of its initial value?

This means [A]/[A]₀ = 63.25/100 = 0.6325

To answer the question we write the repression

0.6325 = e^–kt and solve for t

Taking the natural logarithm of both sides gives ln(0.6325) = –kt = –(3.237 × 10^–2)t

So t = –ln(0.6325)/(3.237 × 10^–2) = 14.15 s

Question 2

How long does it take for the concentration of the reactant in the reaction to fall to 13.75% of its initial value?

Follow the same procedure as above to arrive at t = –ln(0.1375)/(3.237 × 10^–2)

Here I'll leave the calculator work to you...

Hope this helps. Add a comment if you have any questions.