Answer : The total pressure (in atmospheres) in the reaction vessel when equilibrium reached is, 5.02 atm
Explanation :
First we have to calculate the initial pressure of
and ![O_2](https://tex.z-dn.net/?f=O_2)
![P_{SO_2}=\frac{n_{SO_2}RT}{V}](https://tex.z-dn.net/?f=P_%7BSO_2%7D%3D%5Cfrac%7Bn_%7BSO_2%7DRT%7D%7BV%7D)
![P_{SO_2}=\frac{(0.100mol)\times (0.0821L.atm/mol.K)\times (950K)}{2.75L}](https://tex.z-dn.net/?f=P_%7BSO_2%7D%3D%5Cfrac%7B%280.100mol%29%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20%28950K%29%7D%7B2.75L%7D)
![P_{SO_2}=2.84atm](https://tex.z-dn.net/?f=P_%7BSO_2%7D%3D2.84atm)
and,
![P_{O_2}=\frac{n_{O_2}RT}{V}](https://tex.z-dn.net/?f=P_%7BO_2%7D%3D%5Cfrac%7Bn_%7BO_2%7DRT%7D%7BV%7D)
![P_{O_2}=\frac{(0.100mol)\times (0.0821L.atm/mol.K)\times (950K)}{2.75L}](https://tex.z-dn.net/?f=P_%7BO_2%7D%3D%5Cfrac%7B%280.100mol%29%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20%28950K%29%7D%7B2.75L%7D)
![P_{O_2}=2.84atm](https://tex.z-dn.net/?f=P_%7BO_2%7D%3D2.84atm)
The balanced chemical reaction is:
![2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)](https://tex.z-dn.net/?f=2SO_2%28g%29%2BO_2%28g%29%5Crightleftharpoons%202SO_3%28g%29)
Initial pressure 2.84 2.84 0
At eqm. (2.84-2x) (2.84-x) 2x
The expression of
for above reaction follows:
![K_p=\frac{(P_{SO_3})^2}{(P_{SO_2})^2\times P_{O_2}}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%28P_%7BSO_3%7D%29%5E2%7D%7B%28P_%7BSO_2%7D%29%5E2%5Ctimes%20P_%7BO_2%7D%7D)
Putting values in expression 1, we get:
![0.355=\frac{(2x)^2}{(2.84-2x)^2\times (2.84-x)}](https://tex.z-dn.net/?f=0.355%3D%5Cfrac%7B%282x%29%5E2%7D%7B%282.84-2x%29%5E2%5Ctimes%20%282.84-x%29%7D)
x = 0.664
Now we have to calculate the total pressure.
Partial pressure of
= (2x) = 2(0.664) = 1.33 atm
Partial pressure of
= (2.84-2x) = [2.84-2(0.664)] = 1.51 atm
Partial pressure of
= (2.84-x) = [2.84-0.664] = 2.18 atm
Total pressure = Partial pressure of
+ Partial pressure of
+ Partial pressure of ![O_2](https://tex.z-dn.net/?f=O_2)
Total pressure = 1.33 + 1.51 + 2.18
Total pressure = 5.02 atm
Thus, the total pressure (in atmospheres) in the reaction vessel when equilibrium reached is, 5.02 atm