Hydrogen iodide decomposes at 800 K via a second-order process to produce hydrogen and iodine according to the following chemical equation. 2 HI(g) → H₂(g) + I₂(g)
At 800 K it takes 142 seconds for the initial concentration of HI to decrease from 6.75 × 10^-2 M to 3.50 × 10^-2 M.What is the rate constant for the reaction at this temperature?

Question

Quizplus · Accepted Answer

We can use the second-order rate equation,

Rate = k[HI]^2

We are given the initial concentration of HI and the time taken for it to decrease from 6.75×10^-2 to 3.50×10^-2, so we can calculate the change in concentration,

Δ[HI] = (6.75×10^-2 - 3.50×10^-2) = 3.25×10^-2 M

We can also calculate the average concentration of HI over this time period,

[HI]avg = (6.75×10^-2 + 3.50×10^-2)/2 = 5.125×10^-2 M

Now we can plug in these values and solve for the rate constant,

k = Rate/[HI]^2 = Δ[HI]/t/[HI]^2 = (3.25×10^-2)/(142 s)(5.125×10^-2)^2

k = 9.69×10^-12 M^-1 s^-1

So the answer is (B).

Hydrogen Iodide Decomposes at 800 K Via a Second-Order Process