Question

Sulfur dioxide, SO2 is oxidized in the presence of the catalyst vanadium(V) oxide, V2O5. In this reaction 2 mol of SO2 and 1.4 mol of O2 are mixed in 3 dm^3 sealed container and the system is allowed to come to equilibrium. At 700K a conversion rate of 15% is achieved. Calculate the equilibrium constant for this reaction.

Answer 1

The equilibrium constant (K) for a reaction is determined by the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. In this case, the equilibrium constant can be calculated using the given concentrations of SO3, SO2, and O2. The calculated value of K is 0.771.

Step-by-step explanation:

The equilibrium constant (K) for a reaction is determined by the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. In this case, the equilibrium equation is 2SO2(g) + O2(g) -> 2SO3(g).

At equilibrium, the ratio of the concentrations of SO3 to SO2 and O2 is given as 5.0 x 10^-2 M SO3, 3.0 x 10^-2 M SO2, and 3.5 x 10^-7 M O2.

To calculate K, you need to raise the concentrations of the products (SO3) to their stoichiometric coefficients and divide by the concentrations of the reactants (SO2 and O2) raised to their stoichiometric coefficients. K = [SO3]^2 / [SO2]^2 * [O2] = (5.0 x 10^-2)^2 / (3.0 x 10^-2)^2 * (3.5 x 10^-7) = 0.771.