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

2. Le Chatellier’s principle and more... – Nice video- 20 minutes Nice video- 20 minutes – Another good one- 15 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

3. Stress 1. Temperature change this is the ONLY stress that would actually change K c increasing temperature – favors the “cold side”/endothermic/the reaction that needs heat – adding heat is like adding more products to the reaction so therefore it shifts left to counteract stress – K c decreases decreasing temperature – favors the “hot side”/exothermic/the reaction that gives off heat – K c increases  H = + 92 kJ N 2 (g) + 3H 2 (g)   2NH 3 (g)  H = - 92 kJ Haber Process again McGraw Hill Flash animation

4. Stress 2. Pressure change an increase in pressure causes the equilibrium to shift in the direction that has the fewer number of moles – results in a decrease in N 2 and H 2 and an increase in NH 3 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 N 2 and H 2 and an decrease in NH 3 does NOT affect the equilibrium constant K c N 2 (g) + 3H 2 (g)   2NH 3 (g)  H = - 92 kJ Haber Process again McGraw Hill Flash animation

5. Stress 3. Concentration change 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 K c N 2 (g) + 3H 2 (g)   2NH 3 (g)  H = - 92 kJ Haber Process again K c = [NH 3 ] 2 [N 2 ] [H 2 ] 3 McGraw Hill Flash animation

6. Practice Problem Predict the effect of the following changes on the reaction in which SO 3 decomposes to form SO 2 and O 2. 2 SO 3 (g)   2 SO 2 (g) + O 2 (g) H o = 197.78 kJ increasing the temperature of the reaction – shifts right increasing the pressure on the reaction – shifts left adding more O 2 when the reaction is at equilibrium – shifts left removing O 2 from the system when the reaction is at equilibrium – shifts right

7. 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

8. Le Chatelier’s Principle – Summary 8 ChangeEffect on EquilibriumChange in Kc? Increase concentrationShifts to opposite sideNo Decrease concentrationShifts to same sideNo Increase pressureShifts to side with least moles of gas No Decrease pressureShifts to side with most moles of gas No Increase temperatureShifts in endothermic direction Yes Decrease temperatureShifts in exothermic direction Yes Add a catalystNo changeNo