The pollutant SO_2 can be removed using the following equilibrium: CaO_(s) + SO_

Question

The pollutant SO_2 can be removed using the following equilibrium: CaO_(s) + SO_2 (g) CaSO_3 (s) Delta H degree = -274.9 kJ State what would happen to the equilibrium position and the reason of the shift if the following occurred. Some of the CaSO_3 were removed from the system? Some CaO were added to the system? The temperature was increased? The pressure is decreased? Write the equilibrium constant expression (K_c) for this reaction. In order to remove as much SO_2 as possible, what would be good operating conditions for this reaction? (high/low) temperature (high/low) pressure (high/low) concentration of

Explanation / Answer

For the given reaction,

(a) When some of the CaSO3 is removed, according to LeChatellier's principle the equilibrium would shift in the opposite direction to compensate the change and reestablish the equilibrium.

Thus the reaction would shift towards right handside, that is prouduct end and more of CaSO3 would be formed.

(b) If some of CaO is removed, the reaction would shift towards left handside, that is reactant end.

(c) Temperature is increased. As the reaction is exothermic in nature. Increase in temperature would shift the reaction towards left handside, that is reactant end so that the energy provided in the form of heat is reduced and restablish of equilibrium occurs.

(d) P is decreased. The reaction would shift towards right handside that is towards the product end to reduce the total gas molecules.

(e) Kc for the reaction = 1/[SO2]

(f) In order to remove maximum SO2 we would use the following operating conditions,

low Temperature

low Pressure

high concentration of CaO

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