A first-order reaction is 50% completed in 40 minutes.

For a first-order reaction, the integrated rate law is given by:

ln(A_t) = -kt + ln(A_0)

where:

• A_t is the concentration of the reactant at time t
• A_0 is the initial concentration of the reactant
• k is the rate constant
• t is the time

If the reaction is 50% completed, then A_t = 0.5 A_0. Substituting this into the integrated rate law, we get:

ln(0.5 A_0) = -kt + ln(A_0)

ln(0.5) + ln(A_0) = -kt + ln(A_0)

ln(0.5) = -kt

k = -ln(0.5) / t

Given that t = 40 minutes, we can calculate the rate constant k:

k = -ln(0.5) / 40 min

k ≈ 0.0173 min⁻¹

Therefore, the rate constant for this first-order reaction is approximately 0.0173 min⁻¹.