Sulfur Dioxide reacts with oxygen to produce Sulfur trioxide in the presence of

Question

Sulfur Dioxide reacts with oxygen to produce Sulfur trioxide in the presence of a catalyst. In an experiment carried out at 727 degrees Celsius, the partial pressure of SO2 and O2 at equilibrium is found to be 0.721atm and 0.581 atm respectively , with a total pressure of 2.325 atm. what is the value for Kp for this reaction at this temperature? What would be the effect of an increase in Pressure on the amount of SO3 Produced? and what would be the effect of an increase in temperature on the amount of SO3 produced?

Explanation / Answer

2SO2(g) + O2(g) ? 2SO3(g) 2) Change in pressure (by changing volume) - If the pressure of the system is increased by decreasing the volume, the system will try to decrease the pressure. Since the partial pressures of gases are directly proportional to their number of moles, the equilibrium shifts to right. Why? Because, at the right there are 2 moles, at the left there are 4 moles. As the number of moles decreases, the pressure also decreases. - If the pressure of the system is decreased by enlarging the volume, the system will try to increase the pressure. The reverse of the first case will happen. - Same thing occurs when some O2 is removed. 3) Change in temperature - If the temperature of the system is increased, the system will try to decrease this increase by consuming heat. Since the forward reaction is exothermic (?H = ?197 kJ), the reverse reaction will be effective. The equilibrium will shift to left. Endothermic reverse reaction will consume heat. - If the temperature of the system is decreased, the system will try to increase this decrease by poducing heat. Since the forward reaction is exothermic (?H = ?197 kJ), the forward reaction will be effective. The equilibrium will shift to right. Exothermic forward reaction will produce heat

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