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JCHS Honors Chemistry Spring Final S.G (Part 2)

JCHS Honors Chemistry Spring Final S.G (Part 2). CH Williams. CHEMICAL EQUILIBRIUM. Cato Maximilian Guldberg and his brother-in-law Peter Waage developed the Law of Mass Action. Chemical Equilibrium. Reversible Reactions: . A chemical reaction in which the products

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JCHS Honors Chemistry Spring Final S.G (Part 2)

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  1. JCHS Honors Chemistry Spring Final S.G (Part 2) CH Williams

  2. CHEMICAL EQUILIBRIUM Cato Maximilian Guldberg and his brother-in-law Peter Waage developed the Law of Mass Action

  3. Chemical Equilibrium Reversible Reactions: A chemical reaction in which the products can react to re-form the reactants Chemical Equilibrium: When the rate of the forward reaction equals the rate of the reverse reaction and the concentration of products and reactants remains unchanged 2HgO(s)  2Hg(l) + O2(g) Arrows going both directions ( ) indicates equilibrium in a chemical equation

  4. 2NO2(g)  2NO(g) + O2(g)

  5. Law of Mass Action For the reaction: jA + kB lC + mD WhereKis the equilibrium constant, and is unitless

  6. Product Favored Equilibrium Large values for K signify the reaction is “product favored” When equilibrium is achieved, most reactant has been converted to product

  7. Writing an Equilibrium Expression Write the equilibrium expression for the reaction: 2NO2(g)  2NO(g) + O2(g) K =???

  8. Conclusions about Equilibrium Expressions • The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse 2NO2(g)  2NO(g) + O2(g) 2NO(g) + O2(g)  2NO2(g)

  9. Conclusions about Equilibrium Expressions • When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression, raised to the nth power. 2NO2(g)  2NO(g) + O2(g) NO2(g)  NO(g) + ½O2(g)

  10. Significance of the Reaction Quotient • If Q = K, the system is at equilibrium • If Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved • If Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

  11. Solving for Equilibrium Concentration Consider this reaction at some temperature: H2O(g) + CO(g)  H2(g) + CO2(g) K = 2.0 Assume you start with 8 molecules of H2O and 6 molecules of CO. How many molecules of H2O, CO, H2, and CO2 are present at equilibrium? Here, we learn about “ICE”– the most important problem solving technique in the second semester. You will use it for the next 4 chapters!

  12. Solving for Equilibrium Concentration H2O(g) + CO(g)  H2(g) + CO2(g) K = 2.0 Step #1:We write the law of mass action for the reaction:

  13. Solving for Equilibrium Concentration Step #2:We “ICE” the problem, beginning with the Initial concentrations H2O(g) + CO(g)  H2(g) + CO2(g) 8 6 0 0 -x -x +x +x 8-x 6-x x x

  14. Solving for Equilibrium Concentration Step #3:We plug equilibrium concentrations into our equilibrium expression, and solve for x H2O(g) + CO(g)  H2(g) + CO2(g) x = 4

  15. Solving for Equilibrium Concentration Step #4:Substitute x into our equilibrium concentrations to find the actual concentrations H2O(g) + CO(g)  H2(g) + CO2(g) x = 4

  16. Le Chatelier’s Principle

  17. LeChatelier’s Principle Henry Le Chatelier When a system at equilibrium is placed under stress, the system will undergo a change in such a way as to relieve that stress and restore a state of equilibrium.

  18. Le Chatelier Translated: When you take something away from a system at equilibrium, the system shifts in such a way as to replace some what you’ve taken away. When you add something to a system at equilibrium, the system shiftsin such a way as touse up some of what you’ve added.

  19. LeChatelier Example #1 A closed container of ice and water is at equilibrium. Then, the temperature is raised. Ice + Energy  Water The system temporarily shifts to the _______ to restore equilibrium. right

  20. LeChatelier Example #2 A closed container of N2O4 and NO2 is at equilibrium. NO2 is added to the container. N2O4 (g) + Energy  2 NO2(g) The system temporarily shifts to the _______ to restore equilibrium. left

  21. LeChatelier Example #3 A closed container of water and its vapor is at equilibrium. Vapor is removed from the system. water + Energy  vapor The system temporarily shifts to the _______ to restore equilibrium. right

  22. LeChatelier Example #4 A closed container of N2O4 and NO2 is at equilibrium. The pressure is increased. N2O4 (g) + Energy  2 NO2(g) The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation. left

  23. Moles and Formula Mass

  24. The Mole 1 dozen = 12 1 gross = 144 1 ream = 500 1 mole = 6.022 x 1023 There are exactly 12 grams of carbon-12 in one mole of carbon-12.

  25. Avogadro’s Number 6.022 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro (1776-1855). I didn’t discover it. Its just named after me! Amadeo Avogadro

  26. Calculations with Moles:Converting moles to grams How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li 6.94 g Li = g Li 45.1 1 mol Li

  27. Calculations with Moles:Converting grams to moles How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li 1 mol Li 2.62 = mol Li 6.94 g Li

  28. Calculations with Moles:Using Avogadro’s Number How many atoms of lithium are in 3.50 moles of lithium? 6.02 x 1023 atoms 3.50 mol = atoms 2.07 x 1024 1 mol

  29. Calculations with Moles:Using Avogadro’s Number How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li 6.022 x 1023 atoms Li 1 mol Li 6.94 g Li 1 mol Li (18.2)(6.022 x 1023)/6.94 1.58 x 1024 = atoms Li

  30. Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO3. 24.31g + 12.01 g + 3(16.00 g) = 84.32 g

  31. Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO3. From previous slide: 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g 100.00

  32. Formulas Empirical formula: the lowest whole number ratio of atoms in a compound. • molecular formula = (empirical formula)n [n = integer] • molecular formula = C6H6 = (CH)6 • empirical formula = CH Molecular formula: the true number of atoms of each element in the formula of a compound.

  33. Formulas(continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaCl MgCl2 Al2(SO4)3 K2CO3

  34. Formulas(continued) Formulas for molecular compoundsMIGHTbe empirical (lowest whole number ratio). Molecular: C6H12O6 H2O C12H22O11 Empirical: H2O CH2O C12H22O11

  35. Empirical Formula Determination • Base calculation on 100 grams of compound. • Determine moles of each element in 100 grams of compound. • Divide each value of moles by the smallest of the values. • Multiply each number by an integer to obtain all whole numbers.

  36. Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

  37. Empirical Formula Determination(part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

  38. Empirical Formula Determination(part 3) Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 x 2 x 2 3 5 2 C3H5O2 Empirical formula:

  39. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C3H5O2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  40. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 2. Divide the molecular mass by the mass given by the emipirical formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

  41. Finding the Molecular Formula The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3. Multiply the empirical formula by this number to get the molecular formula. 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g (C3H5O2) x 2 = C6H10O4

  42. Solution Formation

  43. Heat of Solution The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.

  44. Heat of Solution The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.

  45. Steps in Solution Formation H2Step 2-Expanding the solvent Overcoming intermolecular forces of the solvent molecules

  46. Steps in Solution Formation H3Step 3 - Interaction of solute and solvent to form the solution

  47. Predicting Solution Formation

  48. Solubility Chart

  49. Ionic Bonding • Electrons are transferred • Electronegativity differences are generally greater than 1.7 • The formation of ionic bonds is always exothermic!

  50. Determination of Ionic Character Electronegativity difference is not the final determination of ionic character Compounds are ionic if they conduct electricity in their molten state

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