2. H 2 O + CO H 2 + CO 2 As a reaction proceeds, the concentration of reactants declines, and the concentration of products increases

3. H 2 O + CO H 2 + CO 2 For reactions with a significant reverse reaction, the decrease in reactant concentration causes a decrease in the rate of the forward reaction. Also, the increase in product concentration causes a decrease in the rate of the forward reaction.

4. H 2 O + CO H 2 + CO 2 At some point, the rates of the forward and reverse reactions become equal and constant. This is called the equilibrium state because the changes in concentration and rate reach a constant, steady state.

5. In the equilibrium state, the concentration of all reactants and products also remains constant. The concentrations are constant but this does not mean the concentrations are equal.

6. The state of equilibrium is a a steady-state system It remains constant unless acted upon by the surroundings.

7. Equilibrium The state of a reaction where the concentrations of all reactants and products remain constant over time The state of a reaction where the forward and rate of the reaction are equal

8. Equilibrium Expression Since the concentration of all reactants and products remains unchanged during equilibrium, the ratio of the product concentrations to the reactants concentrations is a constant. This ratio is used to describe the equilibrium state for a particular reaction at a specified temperature. The ratio is called the equilibrium constant ( K) K c if concentrations are measured in molarity K p if concentrations are measured in pressure

9. Equilibrium constant = K = [ C ] 3 [ D ] [ A ] 2 [ B ] The coefficients of the balance equation are automatically included in the equilibrium expression Equilibrium Expression For the reaction: 2 A + B 3 C + D

10. 1. Pure Solids Since concentration is constant for a solid, a solid reactant or product is left out of the equilibrium equation 2. Pure liquids and Solvents Since the concentration of a pure liquid or solvent is relatively constant, a pure liquid or solvent (reactant or product) is left out of the equilibrium equation

11. CaCO 3 (s) CaO (s) + CO 2 (g) H 2 CO 3 (aq) + H 2 O (l) HCO 3 - (aq) + H + (aq) K = [CO 2 ] K= [HCO 3 - ] [H + ] [H 2 CO 3 ]

12. Dinitrogen tetroxide decomposes on heating to nitrogen dioxide in a reversible reaction. At equilibrium, a two liter flask at 100 ° C contains 0.0090 mol dinitrogen tetroxide and 0.060 mol of nitrogen dioxide. Determine the equilibrium constant for the reaction at 100° C. N 2 O 4 2 NO 2 [N 2 O 4 ] = 0.0045 M [NO 2 ] = 0.030 M note the absence of units for K K = [NO 2 ] 2 = [0.030] 2 = 0.20 [N 2 O 4 ] [0.0045]

13. Tin II oxide reacts with carbon monoxide by the following reaction: SnO 2 (s) + 2 CO (g) Sn (s) + 2 CO 2 (g) Measurements indicate that 10 grams of SnO 2 placed in a chamber filled with CO gas results in a partial pressure of 2.50 atm for carbon monoxide and a partial pressure of 3.12 atm for carbon dioxide at equilibrium. Write the equilibrium expression for the reaction. K p = (P CO2 ) 2 (note that the equilibrium constant is K p (P CO ) 2 since the concentration is in pressure. What is the value of the equilibrium constant K p K p = (P CO2 ) 2 = ( 3.12) 2 = 1.56 (P CO ) 2 (2.50) 2 How would the equilibrium concentrations of CO and CO 2 change if the amount of SnO 2 is increased to 20 g? No change