Respuesta :
Answer : The partial pressure of [tex]SO_3[/tex] is, 88.84 atm
Solution :
The given equilibrium reaction is,
[tex]2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)[/tex]
[tex]\text{Partial pressure of }SO_2[/tex] = 36.9 atm
[tex]\text{Partial pressure of }O_2[/tex] = 16.8 atm
The expression of [tex]K_p[/tex] will be,
[tex]K_p=\frac{(p_{SO_3})^2}{(p_{O_2})(p_{SO_2})^2}[/tex]
Now put all the values in this expression, we get
[tex]0.345=\frac{(p_{SO_3})^2}{(16.8)\times (36.9)^2}[/tex]
[tex](p_{SO_3})=88.84atm[/tex]
Therefore, the partial pressure of [tex]SO_3[/tex] is, 88.84 atm
The study of chemicals and bonds is called chemistry. When the amount of reactant and product gets equal is said to be an equilibrium state.
The correct answer is 88.84 atm.
What is equilibrium constant?
- The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium.
- A state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change.
The balanced reaction is:-
[tex]2SO_2+O_2--->2SO_3[/tex]
The data is given as follows:-
- Partial pressure SO2=36.9
- Partial pressure O2= 16.8
The formula used is as follows:-
[tex]\frac{(P_{So_3})^3}{(PO_2)(P_So_2)^2}[/tex]
Placed all the values and solve them.
Hence, the correct answer is 88.84atm.
For more information about the equilibrium, refer to the link:-
https://brainly.com/question/16759172
