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.

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:

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times p_{O_2}}$

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:

$0.40=\frac{(2x)^2}{(0.52-2x)^2\times (0.52-x)}\\\\x=-1.13,-0.402,0.077$

Neglecting the negative values because partial pressure cannot be negative.

So, x = 0.077

Equilibrium partial pressure of sulfur dioxide = $(0.52-2x)=(0.52-(2\times 0.077))=0.366atm$

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

Equilibrium partial pressure of sulfur trioxide = $2x=(2\times 0.077)=0.154atm$

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