Question

The reaction 2HI → H2 + I2 is second order in [HI] and second order overall. The rate constant of the reaction at 700°C is 1.57 × 10−5 M −1s−1. Suppose you have a sample in which the concentration of HI is 0.75 M. What was the concentration of HI 8 hours earlier?

Answer 1

Answer:

1.135 M.

Explanation:

• For the reaction: 2HI → H₂ + I₂,

The reaction is a second order reaction of HI,so the rate law of the reaction is: Rate = k[HI]².

• To solve this problem, we can use the integral law of second-order reactions:

1/[A] = kt + 1/[A₀],

where, k is the reate constant of the reaction (k = 1.57 x 10⁻⁵ M⁻¹s⁻¹),

t is the time of the reaction (t = 8 hours x 60 x 60 = 28800 s),

[A₀] is the initial concentration of HI ([A₀] = ?? M).

[A] is the remaining concentration of HI after hours ([A₀] = 0.75 M).

∵ 1/[A] = kt + 1/[A₀],

∴ 1/[A₀] = 1/[A] - kt

∴ 1/[A₀] = [1/(0.75 M)] - (1.57 x 10⁻⁵ M⁻¹s⁻¹)(28800 s) = 1.333 M⁻¹ - 0.4522 M⁻¹ = 0.8808 M⁻¹.

∴ [A₀] = 1/(0.0.8808 M⁻¹) = 1.135 M.

So, the concentration of HI 8 hours earlier = 1.135 M.