Respuesta :
Answer: The partial pressure of the [tex]CO,Cl_2\text{ and }COCl_2[/tex] are 0.352 atm, 0.352 atm and 0.408 atm respectively.
Explanation:
We are given:
[tex]K_c=4.63\times 10^{-3}[/tex]
[tex]p_{COCl_2}=0.760atm[/tex]
Relation of [tex]K_p[/tex] with [tex]K_c[/tex] is given by the formula:
[tex]K_p=K_c(RT)^{\Delta ng}[/tex]
Where,
[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = ?
[tex]K_c[/tex] = equilibrium constant in terms of concentration = [tex]4.63\times 10^{-3}[/tex]
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature = [tex]527^oC=527+273=800K[/tex]
[tex]\Delta ng[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}=2-1=1[/tex]
Putting values in above equation, we get:
[tex]K_p=4.63\times 10^{-3}\times (0.0821\times 800)^{1}\\\\K_p=0.304[/tex]
The chemical reaction for the decomposition of phosgene follows the equation:
[tex]COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)[/tex]
At t = 0 0.760 0 0
At [tex]t=t_{eq}[/tex] 0.760-x x x
The expression for [tex]K_p[/tex] for the given reaction follows:
[tex]K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}[/tex]
We are given:
[tex]K_p=0.304\\p_{COCl_2}=0.760-x\\p_{CO}=x\\p_{Cl_2}=x[/tex]
Putting values in above equation, we get:
[tex]0.304=\frac{x\times x}{0.760-x}\\\\x=0.352,-0.656[/tex]
Negative value of 'x' is neglected because partial pressure cannot be negative.
So, the partial pressure for the components at equilibrium are:
[tex]p_{COCl_2}=0.760-0.352=0.408atm\\\\p_{CO}=0.352atm\\\\p_{Cl_2}=0.352atm[/tex]
Hence, the partial pressure of the [tex]CO,Cl_2\text{ and }COCl_2[/tex] are 0.352 atm, 0.352 atm and 0.408 atm respectively.
