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?

1.135 M.

Explanation:

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

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.

sulpher is precipitated

Explanation: