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

But first, a review of Topic 7.1

Feature of Equilibrium State Explanation Equilibrium is dynamic The reaction has not stopped but the forward and reverse reaction are still occurring Equilibrium is achieved in a closed system A closed system prevents exchange of matter with the surroundings The concentrations of reactants and products remain constant They are being produced and degraded at an equal rate The macroscopic properties do not change Color and density are properties that depend upon concentrations Equilibrium can be reached from either direction The same equilibrium mixture will result under the same conditions, regardless of the starting point.

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

aA + bB  cC + dD [C]c[D]d Kc = [A]a[B]b Kc = 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 =

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

Stress 1. Temperature change 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 Kc decreases

decreasing temperature favors the “hot side”/exothermic 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

Stress 2. Pressure change 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

Stress 3. Concentration change McGraw Hill Flash animation Haber Process again N2 (g) + 3H2 (g)  2NH3 (g) DH = - 92 kJ Kc = [NH3]2 [N2] [H2]3 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

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 removing O2 from the system when the reaction is at equilibrium

Le Chatelier’s Principle – Summary Change Effect on Equilibrium Change in Kc? Increase concentration Shifts to opposite side No Decrease concentration Shifts to same side Increase pressure Shifts to side with least moles of gas Decrease pressure Shifts to side with most moles of gas Increase temperature Shifts in endothermic direction Yes Decrease temperature Shifts in exothermic direction Add a catalyst No change

Le Châtelier’s Principle Changes in Concentration Remove Add Add Remove aA + bB cC + dD 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

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 N2 (g) + 3H2 (g) 2NH3 (g) Add NH3 14.5

Le Châtelier’s Principle Changes in Volume and Pressure (Only a factor with gases) A (g) + B (g) C (g) 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

N2(g) + O2(g)  2NO(g) ∆H = +kJ mol-1 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

Position of Equilibrium Effect of: Position of Equilibrium Value of Kc Concentration Changes No change Pressure Changes if reaction involves a change in the number of gas molecules Temperature Changes (depends on if exothermic or endothermic) Catalyst