Question

A first order reaction has a rate constant of 0.0051 min^–1. If we begin with 0.10M concentration of the reactant, what concentration of reactant will remain in solution after 3 hours?

Answer 1

The first order rate equation which connects the rate constant and concentration of reactant with time is:

\(k = \frac{2.303}t \log_e \frac{A_0}{A}\), 

where k is rate constant, t is time, A is initial concentration and A₀ is left over concentration after time t.

Given values,

k = 0.0051 min⁻¹ 

t = 3 hours

= 3 × 60 minutes

A = 0.10 M

As the quantities are in the same units now, we will use the first order rate equation to find the concentration after the given time.

Putting the values,

⇒ \(0.0051 = \frac{2.303}{3 \times 60} \log_e \frac{0.10}A\)

⇒ \(\log_e \frac{0.10}A = \frac{0.0051 \times 180}{2.303}\)

⇒ \(\log_e \frac{0.10}A = 0.3986\)

Taking antilog to the base e on both sides,

⇒ \(\frac{0.10}A = 2.503\)

⇒ \(A = 0.039\) M

Hence the concentration of the reactant after 3 hours is 0.039 M.

Answer 2

For a first order reaction