Question

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?

Answer 1

Rate constant for the reaction at 800K = 0.0968 per Ms.

Explanation:

Given: At 800 K, initial concentration of HI = 6.75 × 10-2 M = 0.0675 M

Final concentration of HI = 3.50 × 10-2 M = 0.035 M

Time t = 142 seconds.

Solution: According to the second order rate law, we know that:

1 / [HI] = kt + 1 / [HI]o

Substituting the relevant values, we get:

1 / 0.035 = k * 142 + 1 / 0.0675

k = 0.0968 per Ms.

So rate constant for the reaction at 800K = 0.0968 per Ms.