A first order reaction takes 20 minutes for 25% decomposition

The question is incomplete. It should specify the percentage of decomposition that occurs in 20 minutes. Let's assume the question is: 'A first-order reaction takes 20 minutes for 25% decomposition. What is the rate constant?'

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

ln(Nt/N0) = -kt

where:

• Nt is the amount of reactant remaining after time t
• N0 is the initial amount of reactant
• k is the rate constant
• t is the time

If 25% of the reactant decomposes, then 75% remains. Therefore, Nt/N0 = 0.75. We are given that t = 20 minutes. We can solve for k:

ln(0.75) = -k(20 minutes)

k = -ln(0.75) / 20 minutes

k ≈ 0.0144 minutes-1

Therefore, the rate constant for this first-order reaction is approximately 0.0144 minutes-1. The half-life (t1/2) can also be calculated using the equation:

t1/2 = ln(2) / k

t1/2 ≈ ln(2) / 0.0144 minutes-1

t1/2 ≈ 48 minutes

This means it takes approximately 48 minutes for half of the reactant to decompose.