Answer:
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
