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Le Chatelier’s principle and more... 7.2.3-7.2.5

Le Chatelier’s principle and more... 7.2.3-7.2.5. But first, a review of Topic 7.1. Catalyst. the same process still has to happen, catalysts just help out by lowering the activation energy

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Le Chatelier’s principle and more... 7.2.3-7.2.5

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  1. Le Chatelier’s principle and more...7.2.3-7.2.5

  2. But first, a review of Topic 7.1

  3. Catalyst • the same process still has to happen, catalysts just help out by lowering the activation energy • increase the RATE of a reaction….and therefore the decrease the time in which equilibrium is reached • they speed up the forward and reverse reactions equally • therefore decreases the time required for the system to achieve equilibrium • less time equals $$$ when making chemicals

  4. [C]c[D]d Kc= [A]a[B]b aA + bB cC + dD • Write the equilibrium expression for the following reaction: • 4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g) • [NO2]4[H2O]6 • [NH3]4 [O2]7 Kc=

  5. Le Chatellier’s principle and more... • Nice video- 20 minutes • Another good one- 15 minutes • states when a system in chemical equilibrium is disturbed by a change, the system shifts in a way that tends to counteract this change of variable • a change imposed on an equilibrium system is called a stress • a stress usually involves a change in the temperature, pressure, or concentration • the equilibrium always responds in such a way so as to counteract the stress

  6. McGraw Hill Flash animation Stress 1. Temperature change Haber Process again DH = + 92 kJ N2(g) + 3H2(g)  2NH3(g) DH = - 92 kJ • this is the ONLY stress that would actually change Kc • increasing temperature • favors the “cold side”/endothermic/the reaction that needs heat • adding heat favors the reaction to the left since it needs +92 kJ • Kcdecreases

  7. DH = + 92 kJ N2(g) + 3H2(g)  2NH3(g) DH = - 92 kJ • decreasing temperature • favors the “hot side”/exothermic • the reaction is already giving off heat it doesn’t need so cooling down is good • Kc increases

  8. McGraw Hill Flash animation Stress 2. Pressure change Haber Process again N2(g) + 3H2(g)  2NH3(g) DH = - 92 kJ • an increase in pressure causes the equilibrium to shift in the direction that has the fewer number of moles • results in a decrease in N2 and H2 and an increase in NH3 • an decrease in pressure causes the equilibrium to shift in the direction that has the most number of moles • results in a an increase in N2 and H2 and an decrease in NH3 • does NOT affect the equilibrium constant Kc

  9. McGraw Hill Flash animation Stress 3. Concentration change Haber Process again N2(g) + 3H2(g)  2NH3(g) DH = - 92 kJ • the equilibrium responds in such a way so as to diminish the increase or equalize the ratio • increasing concentration of reactants shifts the reaction to the right (forward, more product) • increasing concentration of products shifts the reaction to the left (reverse, more reactants) • does NOT affect the equilibrium constant Kc

  10. Practice Problem • Predict the effect of the following changes on the reaction in which SO3 decomposes to form SO2 and O2. 2 SO3(g)  2 SO2 (g) + O2 (g) Ho = 197.78 kJ • increasing the temperature of the reaction • shifts right • increasing the pressure on the reaction • shifts left • adding more O2 when the reaction is at equilibrium • shifts left • removing O2 from the system when the reaction is at equilibrium • shifts right

  11. Le Chatelier’s Principle – Summary

  12. Remove Remove Add Add aA + bBcC+ dD Le Châtelier’s Principle • Changes in Concentration Change Shifts the Equilibrium Increase concentration of product(s) left Decrease concentration of product(s) right Increase concentration of reactant(s) right Decrease concentration of reactant(s) left

  13. N2(g) + 3H2(g)2NH3(g) Add NH3 Le Châtelier’s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position. • Changes in Concentration 14.5

  14. A (g) + B (g) C (g) Le Châtelier’s Principle • Changes in Volume and Pressure • (Only a factor with gases) Change Shifts the Equilibrium Increase pressure Side with fewest moles of gas Decrease pressure Side with most moles of gas Increase volume Side with most moles of gas Decrease volume Side with fewest moles of gas

  15. For each of the following reactions, predict how the equilibrium will shift as the temperature is increase N2(g) + O2(g) 2NO(g) ∆H = +kJ mol-1 2SO2(g) + O2(g)  2SO3(g) ∆H = -kJ mol-1 Right left

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