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Chemical Reactions

Chemical Reactions. Q3U3 Molarity , Solubility, Equilibrium, and Determining Experimental Yield. Solutions and Molarity. Solutions consist of 2 components solute: a substance that is dissolved Solvent: the medium it is dissolved in

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Chemical Reactions

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  1. Chemical Reactions Q3U3 Molarity, Solubility, Equilibrium, and Determining Experimental Yield

  2. Solutions and Molarity • Solutions consist of 2 components • solute: a substance that is dissolved • Solvent: the medium it is dissolved in • Solutions are homogeneous mixtures: constant composition throughout • Solutes do not have to be solids, • They can be gas • CO2 dissolves in water, as does O2 (this is what fish breathe) • They can be liquid • Ethylene glycol (antifreeze) • The most common solvent is water! • The strength of a solution is important and can be described in a quantitative way

  3. Molarity • The Mole is used to describe the concentration of a solution • Or the “strength” of a solution • A 1 Molar solution contains 1 mole of a solute dissolved in enough solvent (usually water) to make 1 liter of solution • Molarity (M) = moles of solute liter of solution One mole of NaCl is 58.5g, if 2 moles of NaCl are dissolved in water to make 1 liter of solution, the solution is a 2M solution This number is very consistent and uniform in all 2 molar solutions of NaCl

  4. Example • Calculate the Molarity of 1.50 L of solution that contains 200.0g of MgCl2 Molarity= (200.0g MgCl2) X (1 molMgCl2) = 1.40 mol/L ( 1.50 L ) (95.3 g MgCl) =1.40 M

  5. Example 2 • Calculate the molarity of a solution that contains 10.0g of sodium hydroxide in 5.00 X 102mL of solution • Molarity = 10.0 g NaOH X 1 mol NaOH X 1000mL 5.0 X 102mL 40.0 g NaOH 1L = 0.500 mol/L = .500 M

  6. Home Work • Pg 109 #’s 1-3

  7. Home Work Answers • a. 1.00 M b. 0.25 M c. 1.21 M • a. 41.6 g b. 421 g c. 71.0 g • a. 1.00 L b. 4.00 L c. 0.737 L

  8. Mole Fraction (X) • Another way to express concentration of solutions • Defined as the moles of a substance divided by the moles of solution (sum of moles of solute and moles of solvent) • X = moles solute (moles solute + moles solvent) The sum of all of the mole fractions in a solution equals 1

  9. Mole Fraction Example • What are the mole fractions of glucose and water in a solution made of 7.59g of glucose, dissolved in water? Molecular mass of glucose is 180.0g, find the moles of glucose. mol glucose= (7.59g glucose) (1 mol glucose)= 0.0422 mol ( 1 ) (180.0 g glucose) The molecular mass of water is 18.0g, find the moles of water. mol water = (125 g water) (1 mol water) = 6.94 mol (18.0 g water)

  10. Mole Fraction Example continued • Now find the mole fraction of glucose Mole fraction = mole solute/(mole solute + mol solvent) =0.0422mol/ (0.0422+6.94)= 0.00605 Now find the mole fraction of water Mol fraction water = 6.94/6.98 =0.994 If you multiply the mole fraction by 100, you get the mole percent!

  11. Now you try these • Calculate the mole fraction of each component in the following 1. 22.5g CH3CH2OH in 1.0 x 102 g H2O H2O = 0.919 CH3CH2OH = 0.0807 2. 39.5 g C6H5CH3 toluene, in 1.5 X 102 g C6H6, benzene C6H5CH3 = 0.183 C6H6 = 0.817

  12. Quiz Calculate the molarity of the following solution 1. 500.0mL with 82.0g Ca(NO3)2 Calculate the mass of solute in the following solutions 2. 250.0mL Na2SO4. 7H2O solution that is 2.0 M 3. 1.50 L of KH2PO4 solution that is 0.24 M How many liters of solution can be made from the following? 4. A 0.10 M solution using 117.0g NaCl 5. A 1.00 M solution using 50.0 g CuSO4 5H2O 6. 0.20 M solution using 200.0g Na2S

  13. Quiz problem solutions 1. Molarity of ---500.0mL with 82.0g Ca(NO3)2 (82.0 g/500.mL) (1mol/164.0g) (1000mL/1L) =1.o M 2. Find mass --250.0mL Na2SO4. 7H2O solution that is 2.0 M (250.0mL/1) (2.0 mol/1L) (268g/1mol) (1L/1000mL) =134g 3. Find mass of -- 1.50 L of KH2PO4 solution that is 0.24 M (1.50 L/1) (0.24 moles/1L) ( 136g/1mol) = 48.96 g KH2PO4 4. liters solution ? -- a 0.10 M solution using 117.0g NaCl (117.0g/1) (1 mol/58.5 g) (1L/0.1 mol)= 20.0L 5. liters?- 1.00 M solution using 50.0 g CuSO4 5H2O (50.0g/1) (1 mol/249.5 g) (1L/1 mol) =0.20 L 6. liters?- 0.20 M solution using 200.0g Na2S (200.g/1) (1 mol/78.0 g) (1 L/0.2 mol) = 12.8 L

  14. Equilibrium Reactions • Previously we have assumed that all reactions have gone to completion ( react until one of the reactants is used up, then the reaction stops) • Reactions go to completion because the elements from the initial reaction are removed in one of the following 3 ways: • Formation of a gas • Formation of a slightly ionized substance • Formation of a precipitate Experimentation has proved that, in some cases, not all of the reactants are converted to the product, no matter how much time is given. These reactions are reversible!

  15. Equilibrium Reactions • Equilibrium reactions do not go to completion • As the reaction proceeds, the reactant concentrations decrease as the product is formed. • However at a certain point, the concentrations level off. • There is no additional product formed, or additional reactant consumed • The concentration of reactants and products no longer change, A state of Chemical equilibrium has been established. • Ex. N2 (g) + 3H2 (g) <----> 2 NH3 (g) + energy A reversible chemical reaction is in equilibrium when : • the rates of the opposing reactions are equal and • overall concentrations remain constant

  16. Le Chatelier’s Principle and Reactants • Le Chatelier’s Principle: If a system in equilibrium is subjected to a disturbance, the equilibrium will shift in an attempt to reduce the disturbance and regain equilibrium • Disturbances or outside influences can consist of the following: • Concentration change • Pressure change • Temperature change • ALSO> If more reactant is added to the system in equilibrium, the reaction shifts right and more product is formed

  17. Le Chatelier’s Principle and Reactants • Concentration Change Example: Ammonia equilibrium equation N2 (g) + 3H2 (g) ----- 2NH3 (g) + energy • Adding N2 (g) disturbs the system, the reaction shifts right • This can be corrected through the consumption of N2, so the reaction shifts right and more NH3 is made. • If H2 (reactant) is removed, the reaction shifts left • The disturbance is relieved by producing more H2 and N2, consuming NH3

  18. Scale /gumdrops demo

  19. Le Chatelier’s Principle and Reactants • Pressure Change Example: Ammonia equilibrium equation N2 (g) + 3H2 (g) ----- 2NH3 (g) + energy NOTE! Pressure Change Only Affects Gaseous equilibrium systems! • If Pressure on reactant gas is increasing, this causes the reaction to shift towards the side with the least volume. • In the above reaction, the shift would be to the right because this would make 4 particles (N2(g) + 3H2) combine to make only 2 particles (NH3) • This would reduce particle collisions, this reduces the pressure, thereby relieving the disturbance

  20. Pressure Demo-Gumdrops

  21. Le Chatelier’s Principle and Reactants • Temperature Change Example: Ammonia equilibrium equation N2 (g) + 3H2 (g) ----- 2NH3 (g) + energy A temperature increase will cause the reaction to shift in the direction of the endothermic reaction • In the example, the left to right reaction is exothermic produces heat • A rise in temp will therefore shift the reaction toward the left (endothermic reaction)! • Using up the heat, corrects the disturbance

  22. Le Chatelier’s Principle and Reactants • The Following disturbances do not cause equilibrium shifts • Catalysts, which speed up the reaction by lowering the activation energy, DO NOT CAUSE AN EQUILIBRIUM SHIFT! • They speed up the reaction but do not increase the amount of product produced • Inhibitors, which slow down reactions, also do not cause a shift. • They cause the reaction to take longer, but do not reduced the amount of product produced.

  23. Practice Equilibrium Problems • For the following gaseous equilibrium reactions, indicate what happens to the equilibrium position when the given disturbance or condition change occurs. (shift left, right or no shift) • Remove NH3 gas b. Decrease pressure N2 + 3H2  2NH3 + energy c. Decrease temperature d. add a catalyst CO2 + H2 + energy  CO + H2O e. Increase SO2 concentration f. increase temperature 2SO2 + O2  2 SO3 + energy g. Increase temperature h. increase CO concentration CO2 + C + energy  2CO

  24. Equilibrium Problems • For the following gaseous equilibrium reactions, indicate what happens to the equilibrium position when the given disturbance or condition change occurs. (shift left, right or no shift) • Decrease pressure j. remove N2O4 N2O4 + energy  2 NO2 k. Increase H2 concentration l. increase pressure H2 + Cl2  2HCl + energy m. Decrease O2 concentration n. add catalyst N2 + O2 + energy  2NO

  25. Answers • A. right • B. left , gas will fill space • C. left, produces heat (exothermic) • D. no change, increases speed but does not change volume • E. right • F. left, toward endothermic • G. right, toward endothermic • H. left, consume excess • I. right, increase volume • J. left • K. right • L. no change, particle number is not changed either way • M. left • N. no change

  26. Solubility • The submicroscopic interactions that occur between water molecules and various solute particles determine the extent to which water is able to dissolve a solute. • Ionic compounds • Like salt in water • Dissociation: the process by which changed particles in an ionic solid separate from one another • Produces ions • Covalent compounds • Sugar • No dissociation, just separation (neutral particles)

  27. Solubility • Both cases: interparticle attractive forces between the solvent and the solute particles, overcome attractive forces between the solute particles • Like dissolves like: dissolving occurs when similarities occur between solvent and solute • Polar and polar • Non-polar and non-polar

  28. Solution Saturation • Saturated solution: a solution that holds the maximum amount of solute per solution, the solution can not dissolve any more solute • Unsaturated solution: The amount of solute dissolved is less than the amount of solute than can be dissolved, the solvent can dissolve more solute • Supersaturated solution: Unstable solutions which contain more dissolved solute than they can usually hold. Conditions were manipulated to get additional solute dissolved, temporarily. • Ex fudge making, more sugar is dissolved by heating the mixture, then slow cooling with constant stirring keeps the sugar trapped in small crystals in the fudge

  29. Solubility • Temperature effects solubility! • The solubility of most SOLID solutes generally increases with increasing temperature • The solubility of GASES always decreases with increased temperature • The amount of increase or decrease depends on the individual solutes

  30. Limiting Reactants • Many reactants continue until one of the reactants is consumed, the reactant used up first is called the limiting reactant. • This is used only in non-reversible reactions! • Limiting reactant problems are solved by comparing the moles of each reactant. • Steps • Write a balanced equation • Change given to moles • Determine moles produced by given quantity, for each • Determine predicted quantities produced using the lesser amount produced (limiting reaction)

  31. Problem using limiting reactants • Example: If 40.0g H3PO4 react with 60.0g MgCO3. calculate the volume of CO2 produced at STP. 2H3PO4 + 3MgCO3 Mg3(PO4)2 + 3CO2 +3H2O Step 2: Change Grams to moles Mol H3PO4 = 40g H3PO4 1mol H3PO4 = 0.408 mol 1 98.0 g H3PO4 Mol MgCO3 = 60.0 g MgCO3 1mol MgCO3 = 0.712 mol 1 84.3g MgCO3

  32. Ex Problem, limiting reactants continued • Step 3 :determine moles of CO2 which will be produced by each reactant(molar ratio from balanced equation) Mol CO2 = 0.408 mol H3PO4 3 mol CO2 = 0.612 mol 1 2 mol H3PO4 Mol CO2 = 0.712 mol MgCO3 3 mol CO2 =0.712 mol 1 3 mol MgCO3 The limiting reactant produces the lesser amount of product- H3PO4

  33. Example Problem, limiting reactants continued again! Step 4 :Use the limiting reactant to complete the problem (remember molar volume is 22.4L/mole at STP) Volume CO2 = 0.612 mol CO2 22.4 L = 13.7 L at STP 1 1 mol CO2 4og of H3Po4 will produce 13.7 L of CO2 at STP

  34. Now you try one • If 20.og of NaOH react with 30.0g of H2SO4 to produce Na2SO4, which reactant is limiting? 2NaOH + H2SO4  Na2SO4 +2H2O 30 g H2SO4 1 mol H2SO4 = 0.30 mol H2SO4 1 98 g H2SO4 20g NaOH 1mol NaOH = 0.5 mol NaOH 1 40.0g NaOH 0.30 mol H2SO4 1 mol Na2SO4 = 0.3 mol Na2SO4 prod 1 1 mol H2SO4 0.5 mol NaOH 1 mol Na2SO4 = 0.25 mol Na2SO4 prod 1 2 mol NaOH NaOH is limiting reactant, 0.25 mol Na2SO4 prod

  35. Try another one! • If 5.00g of copper metal react with a solution containing 20.0g of AgNO3 to produce silver metal, which reactant is limiting? Cu + 2AgNO3  Cu(NO3)2 + 2Ag Answer: AgNO3, 0.118 mol Ag 5.0 g copper  moles 20.0 g AgNO3  moles Moles and mole ratio # of moles produced 1 mol Cu : 2 mol Ag 2 mol AgNO3 : 2 mol Ag Smallest mole amount produced is limiting reactant

  36. Theoretical and Experimental Yield • Remember • The amount of product produced by a chemical reaction, as predicted by stoichiometry is called Theoretical Yield • when actual yield, produced via experimentation, is different than theoretical yield, the efficiency of the reaction is expressed by determining percent yield. Percent yield = (actual yield X 100 (Theoretical yield) This yield can be affected by many different things

  37. Percent Yield Problems Example 1.  • A student conducts a single displacement reaction that produces 2.755 grams of copper.  • Mathematically he determines that 3.150 grams of copper should have been produced.  • Calculate the student's percentage yield.

  38. Example 1 Solution • Solve: actual amount of product: 2.755 g expected amount of product: 3.150 g   actual amount of productpercentage yield = --------------------------------------- x 100                           expected amount of product                       2.755gpercentage yield = --------------- x 100                            3.150g   • percentage yield = 87.4603174 % = 87.46 %

  39. Example 2 Example 2.  • A student completely reacts 5.00g of magnesium with an excess of oxygen to produce magnesium oxide.  • Analysis reveals 8.10 g of magnesium oxide.  • What is the student's percentage yield?

  40. Solution to Example 2 In this problem, you are given the actual amount of the product, but you are not given the expected amount of the product.  • The second mass shown, 5.00g, is the mass of one of the reactants.  • In order to determine the expected amount of the product in this problem, you must begin with a mass-mass problem.

  41. Solution to Example 2 continued • First, write a balanced chemical equation for the reaction: 2Mg(s) + O2(g) ----> 2MgO(s) • Now, label the given and the unknown: given                    unknown       2Mg(s) + O2(g) ----> 2MgO(s)5.00 g                          X g

  42. Solution to Example 2 continued • Change the mass given to moles by dividing by the molar mass of Mg. mass given: 5.00g molar mass of Mg: 24.3 g                    5.00gNumber of moles of Mg = -----------                                           24.3 g/mole Number of moles of Mg = 0.206 moles

  43. Solution to Example 2 continued • Compare the molar ratio between the given and the unknown to determine the number of moles produced. Coefficient of Mg; 2 Coefficient of MgO:2 # of moles of Mg: = 0.206 moles Number of moles of MgO: = ? number of moles of given      number of moles of unknown--------------------------------   =    -------------------------------------- coefficient of given    coefficient of unknown 0.206 moles     X moles -------------- = ------------- 2                     2 Number of moles of MgO produced = 0.206 moles

  44. Solution to Example 2 continued • Now, change the number of moles of MgO produced to the mass by multiplying by the molar mass of MgO. # of moles of MgO = 0.206 molesMolar mass of MgO = 40.3g/mole mass = # of moles x molar mass mass of MgO = 0.206 moles x 40.3 g/mole mass of MgO = 8.30 g

  45. Solution to Example 2 continued Now, you are ready to solve the percentage yield problem. actual mass of MgO produced = 8.10 g expected mass of MgO = 8.30 g actual amount of product percentage yield = expected amount of product X 100                          8.10 g percentage yield = ------------ x 100                            8.30 g percentage yield = 97.6 %

  46. Your Turn; Try This One • A student completely reacts 6.00g of magnesium with an excess of oxygen to produce magnesium oxide.  • Analysis reveals 8.50 g of magnesium oxide.  • What is the student's percentage yield?

  47. Answer • 2Mg(s) + O2(g) ----> 2MgO(s) • Change the mass given to moles mass given: 6.00g molar mass of Mg: 24.3 g 6.00g Number of moles of Mg = ----------- 24.3 g/mole Number of moles of Mg = 0.247 moles Compare the molar ratio: number of moles of given      number of moles of unknown--------------------------------   =    -------------------------------------- coefficient of given    coefficient of unknown 0.247 moles     X moles -------------- = ------------- 2                     2 Number of moles of MgO produced = 0.247 moles mass = # of moles x molar mass mass of MgO = 0.247 moles x 40.3 g/mole mass of MgO = 9.95 g 8.50 g percentage yield = ------------ x 100                            9.95 g percentage yield = 85.4 %

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