Question

A student ran the following reaction in the laboratory at 1080 K: 2SO3(g) 2SO2(g) + O2(g) When she introduced SO3(g) at a pressure of 0.948 atm into a 1.00 L evacuated container, she found the equilibrium partial pressure of SO3(g) to be 0.369 atm. Calculate the equilibrium constant, Kp, she obtained for this reaction. Kp =

Answer 1

0.078

Step-by-step explanation:

The equation is :

`$2SO_3 (g) \ \ \ \Leftrightarrow \ \ \ 2 SO_2(g) \ \ \ + \ \ \ O_2(g)$`

Initial 0.948 ----- ----

Change -2x +2x +x

Final 0.369 2x x

So the total pressure must reman same = 0.948

And the total pressure = partial pressure of all gases

0.948 = ( 0.369 + 2x + x )

0.948 = 0.369 + 3x

`$x=(0.579)/(3)$`

= 0.193 atm

So the partial pressure of
`$SO_2$` = 0.193 x 2

= 0.386 atm

Partial pressure of
`$O_2$` = 0.193 atm

Therefore,

`$k_p=((P_(SO_2))^2(P_(O_2))^)/((P_(SO_3))^2)$`

`$=((0.386)^2(0.193))/(0.369)$`

= 0.078