Phosphine, PH₃, is a colorless, toxic gas that is used in the production of semiconductors as well as in the farming industry. When heated, phosphine decomposes into phosphorus and hydrogen gases.

4 PH₃(g) ⟶ P₄(g) + 6 H₂(g)

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:

Anytime a reaction is of the first order, the integrated law states that

ln[A]t/[A]₀ = - kt

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

          [A]₀ is the initial concentration of A (phosphine also)

          k is a constant,

          t is the time

We also know that for a first order reaction,

          half life is t1/2 =0.693/k, then, k= 0.693/ t 1/2

where t 1/2 is the half-life.

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

Thus, finding k from the half life time, and then solving it for a time t = 70.5 s

k = 0.693 /  35.0 s = 0.0198 s⁻¹

ln [ PH₃ ]t / [ PH₃]₀ = - kt

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

ln (pPH₃ )t / p(PH₃)₀ = - kt

taking inverse log to both sides of the equation:

(pPH₃ )t / p(PH₃)₀  = - kt

then

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