1. Equilibrium All reactions are reversible In labs, many products of chem.reactions can be directly reacted to form the reactants Sometimes the conditions for the reaction to occur are similar for the forward and reverse reactions, other times they are different –Ex. Fe 3 O 4 + 4H 2  3Fe + 4H 2 O (l) 3Fe + 4H 2 O (g)  Fe 3 O 4 + 4H 2 If steam is not allowed to escape

2. Both reactions are taking place simultaneously at different rates Eventually the two rates are equal Chemical Equilibrium Write one equation with a double arrow

3. Characteristics State of balance between opposing changes can be physical Vapor-liquid Changes can be chemical The original changes before an equilibrium exists are observable Changes in concentration Temperature State

4. Definite relationship between concentrations of reactants and products when reversible reaction is in Equilibrium Example: H 2 (g) + I 2(g)  2HI (g) at 490 0 C [H 2 ]= 1mol/dm 3 [I 2 ]= 1mol/dm 3  When H and I come into contact, reaction starts  Equil.is achieved when reaction rate = product rate

5. The relationship between Reactants and Products is expressed as K eq and can be expressed as: K c(eq) = [Products] [Reactants] Coefficients in the balanced equation are used as superscripts in the equation The Equilibrium Constant (K c ) will not change in a closed system It is affected by changes in temperature

7. What does the K c value determine? Size of value determines the extent a reaction proceeds to the right hand side of a reversible reaction before an equilibrium is achieved A constant value <1 indicates little product forming Constant >1 shows product formation favored The higher the value, the greater the amt.of product

8. Reactions are sensitive to changes in conc., temp. and press. A change in one of these places stress on the system Stress causes the equil. to Shift A new equil. is eventually achieved with the new factors LeChatlier’s Principle

9. LeChatelier’s Principle –System in equilibrium subjected to stress( change in con., temp. or press.) the equilibrium will shift in the direction that tends to counteract the effect of the stress

10. Concentration changes A + B  C + D Add A more AB collisions Forward reaction faster Equil. No longer exists B consumed and its conc. Decreases the reaction slows C + D conc. Increases and its collisions increase and its reaction begins to speed up A new equilibrium is established

11. Temperature changes endo N 2(g) + 3H 2(g)  2NH 3(g) + 92kJ exo Haber Process Every equilibrium has an endo and an exo component  temp. stress countered by favoring the endothermic reaction since it absorbs heat  temp. stress countered by favoring the exothermic reaction as heat is given off

12. Pressure changes Applies mostly to reactions with gases endo N 2(g) + 3H 2(g)  2NH 3(g) + 92kJ exo Haber Process Increase pressure- favors side with fewer moles ( # of particles) of substance Decrease pressure favors side with more moles of substance

13. Calculate an equilibrium constant K Don’t always know conc. of all substances Knowing conc. Of one, can stoichiometrically determine others 1. Tabulate known initial and equil.conc. for all species in the equilibrium 2. For species with known conc. of initial and equilibrium conc., calculate change in conc. as reaction approaches equilibrium 3. Use stoich to determine conc. of all others 4. From init. and changes in conc. Calculate the equilibrium concs. Evaluate K c

14. Example: a mixture of 5.000x10 -3 mol H 2 and 1.000x10 -2 mol I 2 is placed in a 5.000L container at 448 o C and allowed to come to equilibrium. Analysis of the equil. mixture shows the conc. of HI is 1.87x10 -3 M. Calculate K c at 448 o C for the reaction. First: Calculate concentrations of all reactants and products

15. [H 2 ] init = 5.000x10 -3 mol = 1.000x10 -3 M 5.000L [I 2 ] init = 1.000x10 -2 mol = 2.00x10 -3 M 5.000L Place above values into a ICE table I initial conc. C change in conc. E equilibrium conc.

16. H 2(g) + I 2(g)  2HI (g) Initial 1.000X10 -3 M 2.00x10 -3 M 0M Change Equilibrium 1.87x10 -3 M

17. Second: calculate change conc. of HI using initial and equil. values Change is +1.87x10 -3 M Third: use stoichiometry to calculate changes in other substances (1.87x10 -3 mol HI)(1 mol H 2 ) = L 2 mol HI 0.935x10 -3 mol H 2 /L The value for I 2 is the same as H 2

18. H 2(g) + I 2(g)  2HI (g) Initial 1.000x10 -3 M 2.000x10 -3 M 0M Change -0.935x10 -3 M -0.935x10 -3 M +1.87x10 -3 M Equilibrium 1.87x10 -3 M

19. Fourth: calculate equil. conc. using initial conc. and changes Equilibrium conc. H 2 is init. minus consumed [H 2 ] = 1.000x10 -3 M – 0.935x10 -3 M = 0.065x10 -3 M [I 2 ] = 2.00x10 -3 M – 0.935x10 -3 M = 1.065x10 -3 M

20. H 2(g) + I 2(g)  2HI (g) Initial 1.000x10 -3 M 2.000x10 -3 M 0M Change -0.935x10 -3 M -0.935x10 -3 M +1.87x10 -3 M Equilibrium 0.065x10-3 M 1.065x10-3 M 1.87x10 -3 M Finally: now know equilibrium concentrations of reactants and products Use equilibrium expression to calculate K c K c = [HI] 2 = (1.87x10 -3 ) 2 = 51 [H 2 ][I 2 ] (0.065x10 -3 )(1.065x10 -3 )

21. For solutions that have solutes dissolved in water, the same principle can be applied K sp is the solubility constant K sp = [Products] Example: MgCl 2  Mg +2 (aq) + 2Cl -1 (aq) K sp = [Mg +2] [Cl -1 ] 2

22. K of acids and bases K a K b K a = [H + ][cation] = [H + ] 2 K b = [anion][OH - ] = [OH - ] 2 Strong acids and bases dissociate completely into H + or OH - and use above equations Weak acids and bases do not and are similar to equilibrium equations K a or K b = [products] [reactants] use ICE chart