This decomposition is first order with respect to phosphine, and has a half‑life of 35.0 s at 953 K. Calculate the partial pressure of hydrogen gas that is present after 70.5 s if a 2.20 L vessel containing 2.29 atm of phosphine gas is heated to 953 K.

0.57 atm

Explanation:

When a a reaction is first order, we have from calculus the following relation:

ln[A]t/[A]₀ = - kt

where [A]t is the concentration of A (phosphine in this case) after a time, t

[A]₀ is the initial concentration of A

k is the rate constant, and

t is the time

We also know that for a first order reaction

k = 0.693 / t 1/2

wnere t 1/2 is the half-life.

This equation is derived for the case when A]t / = 1/2 x [A]₀ which occurs at the half-life.

Thus, lets first find k from the half life time, and then solve for t = 70.5 s

k = 0.693 / 35.0 s = 0.0198 s⁻¹

ln [ PH₃ ]t / [ PH₃]₀ = - kt

from the ideal gas law we know pV = nRT, so the volumes cancel:

ln (pPH₃) t / p (PH₃) ₀ = - kt

taking inverse log to both sides of the equation:

(pPH₃) t / p (PH₃) ₀ = - kt

thus:

(pPH₃) t = 2.29 atm x e^ ( - 0.0198 s⁻¹ x 70.5 s) = 0.57 atm