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 mol L^-1 to 3.50 × 10^-2 mol L^-1.What is the rate constant for the reaction at this temperature?

Question

Quizplus · Accepted Answer

First, we need to determine the order of the reaction. We can do this using the given rate law:

rate = k[HI]^[2]

where k is the rate constant and [HI] is the concentration of hydrogen iodide.

We can calculate the rate constant by rearranging the above equation:

k = rate / [HI]^[2]

Plugging in the given values:

rate = (6.75 × 10^-12 mol s^-1 - 3.50 × 10^-12 mol s^-1) / 142 s = 2.06 × 10^-14 mol L^-1 s^-1

[HI] = 6.75 × 10^-12 mol L^-1

Substituting into the equation for k:

k = (2.06 × 10^-14 mol L^-1 s^-1) / (6.75 × 10^-12 mol L^-1)^[2] = 9.69 × 10^-2 L mol^-1 s^-1

Therefore, the correct answer is B.

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