Respuesta :
Answer:
the Kp=0.171 atm
Explanation:
For the reaction
COCl₂(g) → CO(g) + Cl₂(g)
since the temperature remains constant , the Kp will remain constant .
Kp = pCO*pCl₂/ pCOCl₂
since for every mole that is converted , a mole of CO(g) and a mole of Cl₂(g) are generated , then
denoting 1 as COCl₂ , 2 as CO , 3 as Cl₂ , 0 as initial state and X as conversion , then the moles of each component will be given by
n₁= n₀*(1-X) , n₂ = n₀*X , n₃ = n₀*X
the final number of total moles will be
nf = n₁ + n₂ + n₃ = n₀*(1-X) + n₀*X + n₀*X = n₀ + n₀*X = n₀ ( 1+X)
then assuming ideal gas behaviour and that the container is rigid (V=constant)
P₀*V= n₀*R*T and Pf*V= nf*R*T
dividing both equations
Pf/ P₀ = n₀*(1+X) / n₀ = (1+X)
X = 1- Pf/ P₀
replacing values
X = 1- Pf/ P₀ = 1- 0.351 atm/0.822 atm = 0.573
thus using Dalton's law
p₁ = Pf*x₁ = Pf*n₀*(1-X)/[n₀ ( 1+X)} =Pf*(1-X)/(1+X)
p₂= Pf*x₂ = Pf*n₀*X/[n₀ ( 1+X)} =Pf*X/(1+X)
p₃= Pf*x₃ = Pf*n₀*X/[n₀( 1+X)} =Pf*X/(1+X)
then the equilibrium constant will be
Kp= p₂*p₃/p₁ = Pf*X²/[(1-X)*(1+X)] = Pf*X²/(1- X²)
replacing values
Kp= Pf*X²/(1- X²) = 0.351 atm * 0.573²/(1-0.573²) = 0.171 atm
