Respuesta :
Answer: [tex]1.58\times 10^{-3}atm[/tex]
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
The given balanced equilibrium reaction is,
[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex] The expression for equilibrium constant for this reaction will be,
[tex]K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2(p_{O_2})}[/tex]
Now put all the given values in this expression, we get :
[tex]0.345=\frac{(p_{SO_3})^2}{(0.125)^2(0.470)}[/tex]
[tex]p_{SO_3}=1.58\times 10^{-3}atm[/tex]
Thus the equilibrium partial pressure of [tex]SO_3[/tex] in the mixture is [tex]1.58\times 10^{-3}atm[/tex]
Considering the definition of chemical equilibrium and equilibrium constant Kp, the equilibrium partial pressure of SO₃ in the mixture is 0.0503 atm.
Chemical equilibrium
Chemical equilibrium is a state of chemical reaction in which no changes are observed over time, despite the fact that the substances present continue to react with each other.
That is, chemical equilibrium is established when two opposite reactions take place simultaneously at the same rate.
Equilibrium constant Kp
The equilibrium constant, Kp, describes the relationship between the concentrations of products and reactants at equilibrium in terms of partial pressures.
For a generic gas phase reaction aA + bB ↔ cC + dD, the expression for Kp is
[tex]Kp=\frac{(P_{C} )^{c}(P_{D}) ^{d} }{(P_{A} )^{a}(P_{B}) ^{b}}[/tex]
In a system in equilibrium, the constant Kp is the quotient of the product of partial pressures of the resulting substances, between the product of the partial pressures of the reacting substances, each of them raised to a power equal to the stoichiometric coefficient involved in the reaction. reaction. Kp is a constant value for each temperature.
Equilibrium partial pressure of SO₃
In this case, the balanced reaction is
2 SO₂(g) + O₂(g) → 2 SO₃(g)
The expression for Kp is
[tex]Kp=\frac{(P_{SO_{3} } )^{2} }{(P_{SO_{2}} )^{2}(P_{O_{2}})}[/tex]
You know:
- Kp= 0.345
- [tex]P_{SO_{3}[/tex]= ?
- [tex]P_{SO_{2}}[/tex]= 0.125 atm
- [tex]P_{O_{2}}[/tex]= 0.47 atm
Replacing in the expression for Kp:
[tex]0.345=\frac{(P_{SO_{3} } )^{2} }{(0.125 atm )^{2}(0.470 atm)}[/tex]
Solving:
[tex](P_{SO_{3} } )^{2}=[/tex]0.345 ×(0.125 atm)² ×0.47 atm
[tex](P_{SO_{3} } )^{2}=[/tex] 2.53×10⁻³ atm³
[tex]P_{SO_{3} } =[/tex] √2.53×10⁻³ atm³
[tex]P_{SO_{3} } =[/tex] 0.0503 atm
The equilibrium partial pressure of SO₃ in the mixture is 0.0503 atm.
Learn more about equilibrium constant Kp:
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