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Unit 10: Chemical Equilibria

Unit 10: Chemical Equilibria. GPS: SC2. Students will relate how the Law of Conservation of Matter is used to determine chemical composition in compounds and chemical reactions. f. Explain the role of equilibrium in chemical reactions.

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Unit 10: Chemical Equilibria

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  1. Unit 10: Chemical Equilibria GPS: SC2. Students will relate how the Law of Conservation of Matter is used to determine chemical composition in compounds and chemical reactions. f. Explain the role of equilibrium in chemical reactions. SC5. Students will understand that the rate at which a chemical reaction occurs can be affected by changing concentration, temperature, or pressure and the addition of a catalyst. a. Demonstrate the effects of changing concentration, temperature, and pressure on chemical reactions. b. Investigate the effects of a catalyst on chemical reactions and apply it to everyday examples.

  2. Equilibrium • EQ: What’s “equal” about equilibrium? • Does it mean that the amounts of reactants and products are equal? • It simply means that there is a balance of materials in a chemical reaction

  3. Equilibrium At Equilibria, the [R] & [P] are constant; but doesn’t mean that [R] = [P]. Equlibrium is a state of action, not inaction ex: arm wrestling match where nobody is winning… there are opposing forces in balance although we don’t see anyone’s arms moving

  4. Equilibrium • Reversible Reaction: one that can occur in both the forward and the reverse directions. N2 + 3 H2 ↔ 2 NH3 • Chemical Equilibrium: a state in which the forward and reverse reactions balance each other because they take place at equal rates. • Equilibrium: opposing forces are in balance.

  5. Chemical Equilibrium • Law of Chemical Equilibrium: at a given temperature, a chemical system may reach a state in which a particular ratio of [R] & [P] has a constant value. • can be expressed w/ a mathematical value, Keq. It is temperature dependent (it will remain constant as long as the temperature does not change.) • Equilibrium constant: Keq, is a numerical value of the ratio of [R] & [P]. • It has no units. It does, however, vary with temperature..... and it tells us which direction the reaction will shift if it is <1, =1, or >1.

  6. How do we interpret Keq, the equilibrium constant? • General equation: aA + bB → cC + dD • Small letters = coefficients • Keq = [C]c[D]d = [product]x [A]a[B]b [reactant]x • [ ] = molar concentration in mol/ L

  7. Keq > 1 : more products at equilibrium • Keq < 1 : more reactants at equilibrium • What if Keq = 1? • Both amounts are exactly the same. • Perfectly balanced equilibrium. • N2 (g) + 3 H2(g) ↔ 2 NH3(g) • [NH3]2 = [product]x [N2][H2]3 [reactant]x

  8. Calculating Keq • If the molar concentrations are given for each R & P, then the value of Keqcan be determine. Just plug in the [ ]’s of each. • N2 (g) + 3 H2(g) ↔ 2 NH3(g) ↓ ↓ ↓ • 0.533M 1.600 M 0.933 M • Keq= [NH3]2 = [0.933M]2 = 0.399 [N2][H2]3 [0.533M][1.600M]3

  9. Calculating Equilibrium Concentrations of R & P • If given keq and [ ]s of all R & P except 1, you can determine the value of the unknown. Just plug in the numbers into the equation and solve for unknown. • 2 H2S(g) ↔ 2 H2 + S2(g) : Keq= 2.27 x 10-3 ↓ ↓ ↓ • 0.184M ? 0.0540M • Keq = [H2]2[S2] [H2S]2 • 2.27 x 10-3=[H2]2 [0.0540M] = [0.184M]

  10. Heterogenous Equilibria When reactants and products are in different phases, pay attention! Pure solids and pure liquids do NOT affect the equilibrium constant at all and are NOT included in the equilibrium expression for finding Keq. Ex: For the reaction PCl5 (s) PCl3 (l) + Cl2 (g) the expression for Keq = [Cl2] Now write the Keq expressions for these examples: 1) As4O6 (s) + 6C (s) As4 (g) + 6CO(g) 2) P4 (s) + 6Cl2 (g)  4PCl3 (l) 3) H2 (g) + I2 (s) 2Hl (g)

  11. LeChatelier’s Principle • LeChatelier’s Principle: when stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress. • Stress is any kind of change in a system at equilibrium that upsets the equilibrium. • There are 3 things that can be manipulated: • Concentration • Volume / Pressure • Temperature

  12. Changes in Concentration • Think of the stress like a seesaw. As one thing is added or removed, it makes the seesaw unbalanced. What must be done in each case to regain that balance? • In the same manner, we can know what direction the reversible reaction will go in (shift).

  13. Changes in Concentration • CO + 3H2 CH4 + H2O If more CO is added, which direction will the seesaw (equilibrium) lean? • It leans to the left, so the reaction must go to the right to compensate and balance the seesaw (equilibrium) out again. • If more CO is added (it is on left side), the reaction shifts to the right to compensate and balance the boat out again (back at equilibrium). • What happens if H2O is removed? • How can we shift the reaction to the left?

  14. Changes in Volume • CO + 3H2 CH4 + H2O • How might changing the volume/pressure affect a reaction? • ↑ V= ↓ P & ↓V = ↑ P • If the pressure increased by decreasing the volume, the molecules would be closer together and would have a better chance of “seeing” each other to react. We would get more product, and the reaction shifts to the right. • When the volume is decreased, the reaction will shift to the less crowded side (has the smallest total # of moles).

  15. Changes in Volume • Notice that although we can shift a reaction in one direction, it is only a temporary change. The system will eventually reach a new equilibrium and balance out again. • Changing the volume/pressure will only shift the direction of the reaction IF the moles of reactants ≠ moles of products! Examples: • Which way would the reaction shift if the volume was decreased? • Remember to compare the TOTAL # moles on each side (look at coefficients) 1) CO(g) + 3H2(g) CH4(g) + H2O(g) 2) 2SO2 + O2  2SO3 3) H2 + Cl2 2HCl 4) 2NOBr  2NO + Br2

  16. Changes in Temperature • Changing the concentration, volume and pressure do not change Keq. It only shifts the position of the equilibrium temporarily. • A change in temperature not only affects equilibrium position, but it is the only thing that can also change Keq.

  17. Changes in Temperature • Virtually all reactions are either endothermic (absorb heat) or exothermic (release heat). • The forward reaction is exothermic, so heat is considered a product. • When heat is a reactant, the reaction is endothermic. (heat goes in / gets used up) • When heat is a product, the reaction is exothermic. (heat goes out / is produced) • Is the reaction endothermic or exothermic? • 4HCl(g) + O2(g) 2H2O(l) + 2Cl2(g) + heat • Changing the temperature will also change the Keq. • Changing the temperature affects the equilibrium & may be observed by a color change.

  18. Catalysts HOMEWORK: What is a catalyst? What effect does a catalyst have on a reaction rate? What effect does a catalyst have on equilibrium? What is an example of a natural catalyst (found in our bodies)?

  19. Biological Equilibrium • Hgb + O2  Hgb(O2) • What happens when at a higher altitude we have an increase in pressure AND a lower O2 concentration in our body? • After being in the mountains for a while, you get used to the altitude and don’t feel the effects anymore. Why?

  20. Practice Problems How does a system at equilibrium respond to stress? What factors are considered to be stresses on an equilibrium system? How would each of the changes affect the equilibrium of this reaction: N2 + 3H2 2NH3 a- Removing hydrogen from the system b- adding more NH3 c- adding more hydrogen

  21. Practice Problems, continued Would you need to raise or lower the temperature to get the wanted result for the following reaction: C2H2 + H2O  CH3CHO + heat a- increase the amount of CH3CHO b- decrease the amount of C2H2 c- increase the amount of water

  22. Solubility Equilibria • Most ionic compounds are highly soluble in water: • NaCl  Na+ + Cl- • Some are only slightly soluble: • BaSO4  Ba+2 + SO4-2 • Although not much dissolves, the reaction eventually reaches equilibrium and the concentration of ions is very small

  23. Solubility Equilibria • The Keq is no longer useful to us and we use a new constant, Ksp. • Ksp : solubility product constant • Ksp for BaSO4 is 1.10 x 10-10 • The Ksp values help us to know how soluble something is

  24. Solubility Equilibria • Ksp can also be used to find out how much of a substance will dissolve into solution • Ex: AgI  Ag+ + I- • Ksp = s2 • s (solubility) = 9.2 x 10-9 mol/L

  25. Solubility Equilibria • We can also use Ksp to find the concentration of ions in a saturated solution • Ex: Mg(OH)2  Mg+2 + 2OH- • Ksp = [Mg+2] + [OH-]2 = 5.6 x 10-12

  26. Common Ion Effect • The solubility of PbCrO4 in water is 4.8 x 10-7 mol/L yet it will not dissolve in a 0.1M solution of K2CrO4. Why? • Ksp = [Pb+2] [CrO4-2] = 2.3 x 10-13 • If the concentration of one ion increases, then the other one must decrease since Ksp is a constant. • Both PbCrO4 and K2CrO4 contribute CrO4-2 ions to solution, a common ion.

  27. Common Ion Effect • Common Ion: an ion common to two or more ionic compounds • Common Ion Effect: when the solubility of a substance is lowered due to the presence of a common ion

  28. When a solution of Pb(NO3)2 is added to a solution of PbCrO4, the PbCrO4 precipitates. • This follows LeChatlier’s Principle: • PbCrO4(g) Pb+2(aq) + CrO4-2(aq) • Adding Pb+2 and NO3-1 to the solution increases the Pb+2 concentration, shifting the reaction to the left.

  29. REVIEW How would each of the following changes affect the equilibrium position (left, right or the same)? CO(g) + 2H2(g)  CH3OH(g) + heat a) adding more CO b) cooling the system c) adding a catalyst d) removing CH3OH e) decreasing the volume

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