Sulfur dioxide and oxygen in the mole ratio 2:1 were mixed at a constant temperature of 1110K and a constant pressure of 9atm in the presence of a catalyst. At equilibrium, one-third of the sulfur dioxide had been converted into sulfur dioxide. Calculate the equilibrium constant Kp for this reaction under these conditions.
Originally posted by LastRide7:Sulfur dioxide and oxygen in the mole ratio 2:1 were mixed at a constant temperature of 1110K and a constant pressure of 9atm in the presence of a catalyst. At equilibrium, one-third of the sulfur dioxide had been converted into sulfur trioxide. Calculate the equilibrium constant Kp for this reaction under these conditions.
Since moles and partial pressures are directly proportional (when temperature and volume remain constant), hence Initial partial pressures of SO2, O2 and SO3 are (2/3)(9)atm, (1/3)(9)atm and 0 atm respectively.
Since at equilibrium, one-third of the sulfur dioxide had been converted into sulfur trioxide, Change will be -(1/3)(2/3)(9)atm for sulfur dioxide and applying stoichiometry you can determine the corresponding partial pressure Changes for oxygen and sulfur trioxide.
Accordingly, you can determine the Equilibrium partial pressures for all 3 species, and consequently the Kp value.