Question

At 920 K, Kp = 0.40 for the following reaction. 2 SO2(g) + O2(g) equilibrium reaction arrow 2 SO3(g) Calculate the equilibrium partial pressures of SO2, O2, and SO3 produced from an initial mixture in which the partial pressures of SO2 and O2 = 0.52 atm and the partial pressure of SO3 = 0 (exactly).

Answer 1

Answer: The equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.

Step-by-step explanation:

We are given:

Initial partial pressure of sulfur dioxide = 0.52 atm

Initial partial pressure of oxygen = 0.52 atm

Initial partial pressure of sulfur trioxide = 0 atm

For the given chemical reaction:

`2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)`

Initial: 0.52 0.52

At eqllm: 0.52-2x 0.52-x 2x

The expression of
`K_p` for above equation follows:

`K_p=((p_(SO_3))^2)/((p_(SO_2))^2* p_(O_2))`

We are given:

`K_p=0.40\\\\p_(SO_3)=2x\\\\p_(SO_2)=0.52-2x\\\\p_(O_2)=0.52-x`

Putting values in above equation, we get:

`0.40=((2x)^2)/((0.52-2x)^2* (0.52-x))\\\\x=-1.13,-0.402,0.077`

Neglecting the negative values because partial pressure cannot be negative.

So, x = 0.077

Equilibrium partial pressure of sulfur dioxide =
`(0.52-2x)=(0.52-(2* 0.077))=0.366atm`

Equilibrium partial pressure of oxygen =
`(0.52-x)=(0.52-0.077)=0.443atm`

Equilibrium partial pressure of sulfur trioxide =
`2x=(2* 0.077)=0.154atm`

Hence, the equilibrium partial pressure of sulfur dioxide, oxygen and sulfur trioxide is 0.366 atm, 0.443 atm and 0.154 atm respectively.