Consider the following equation at equilibrium... 2PbS(s) + 3O2(g) <==> 2PbO(s)
ID: 624361 • Letter: C
Question
Consider the following equation at equilibrium... 2PbS(s) + 3O2(g) <==> 2PbO(s) + 2SO2(g) ... DELTA (H) = -827.4 kJ ... 1) Would the equilibrium yield of PbO increase, decrease, or remain unchanged? EXPLAIN... ... ... 2) Would the value of the equilibrium constant increase, decrease, or remain unchanged in the following situations? EXPLAIN... A) Increasing P(O2) ... B) Adding SO2 to the system ... C) Decreasing the temperature... D) Increasing the volume... E) Increasing the total pressure by adding argon... F) Adding a catalyst...Explanation / Answer
For the first situation you can write it as increasing the pressure of O2 so the equilibrium constant = K = p(S02)^2/p(02)^3 and as equilibrium constant is a constant increasing the pressure of O2 will decrease the value of K so inorder to keep K constant value of p(SO2) should be increased so forward reaction will take place implying yield of PbO increases So in the same way adding SO2 to the system increases the value of K so back ward reaction takes place implying the yield of PbO decreases. As the reaction is exothermic ie heat is given out (This is known through the delta H value) so decreasing the temperature the decreases the K value so inorder to keep it constant forward reaction ie yield increases And increasing the volume of the container makes the equilibrium shifts in the direction of increasing number of moles of gas so back ward reaction so yield decreases and remember volume effect is only seen here as gas molecules are present addition of inert gas like argon doest not effect the equilibrium so no change under constant volume and addition of catalyst doesnt change the yield it just increases the rate of reaction it has nothing to do with the yield So now coming to the value of equilibrium constant K no change with increasing or decreasing the pressure of O2 or SO2 to system and as the reaction is exothermic and as per Vantoff equation value of K increases with decrease in temperature And the pressure difference has no effect and volume effect too and catalyst too as it increases rate of reaction just equilibrium is obtained faster. P S in the equilibrium constant the active masses of solids is taken as one so i left PbS and PbO