The reaction: 2 HI → H₂ + I₂,is second order and the rate constant at 800 K is 9.70 × 10^-2 M^-1 s^-1.How long will it take for 8.00 × 10^-2 mol/L of HI to decrease to one-fourth of its initial concentration?

Question

Quizplus · Accepted Answer

The reaction is second order, so we can use the following integrated rate law:

1/[HI]t - 1/[HI]0 = kt

where [HI]t is the concentration of HI at time t, [HI]0 is the initial concentration, k is the rate constant, and t is the time.

We want to find the time it takes for [HI]t to decrease to one-fourth of its initial concentration, so we can write:

[HI]t = [HI]0/4

1/[HI]t = 4/[HI]0

Substituting into the integrated rate law, we get:

4/[HI]0 - 1/[HI]0 = kt

3/[HI]0 = kt

t = 3/[HI]0k

Plugging in the values given in the problem, we get:

t = 3/(8.00 × 10-2 mol/L)(9.70 × 10-2 M-1s-1)

t = 387 s

Therefore, the answer is (C) 387 s.

The Reaction: 2 HI → H2 + I2,is Second Order

The Reaction: 2 HI → H₂ + I₂,is Second Order