Question

At 900 K the following reaction has Kp=0.345: 2SO2(g)+O2(g)ââ2SO3(g) In an equilibrium mixture the partial pressures of SO2 and O2 are 0.125 atm and 0.470 atm , respectively.
What is the equilibrium partial pressure of SO3 in the mixture?

Answer 1

Answer: $1.58\times 10^{-3}atm$

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

               $2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$       The expression for equilibrium constant for this reaction will be,

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

Now put all the given values in this expression, we get :

$0.345=\frac{(p_{SO_3})^2}{(0.125)^2(0.470)}$

$p_{SO_3}=1.58\times 10^{-3}atm$

Thus the equilibrium partial pressure of $SO_3$ in the mixture is $1.58\times 10^{-3}atm$