Respuesta :
Answer: [tex]p_{SO_2}=0.017atm[/tex]
Explanation:
We are given:
[tex]K_c=1.7\times 10^8[/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
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature = [tex]600K[/tex]
[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]
[tex]\Delta ng[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}=2-3=-1[/tex]
Putting values in above equation, we get:
[tex]K_p=1.7\times 10^8\times (0.0821\times 600)^{-1}\\\\K_p=3.4\times 10^6[/tex]
The chemical reaction follows the equation:
[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]
The expression for [tex]K_p[/tex] for the given reaction follows:
[tex]K_p=\frac{(p_{SO_3})^2}{ p_{O_2}\times {(p_{SO_2})^2}}[/tex]
We are given:
[tex]p_{SO_3}=300atm[/tex] [tex]p_{O_2}=100atm[/tex]
Putting values in above equation, we get:
[tex]3.4\times 10^6=\frac{(300)^2}{100\times {(p_{SO_2})^2}}[/tex]
[tex]p_{SO_2}=0.017atm[/tex]
Hence, the partial pressure of the [tex]SO_2[/tex] at equilibrium is 0.017 atm.
