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Topic 7 Chemical Reactions Review 7 Dec 2017 Agenda FR Sample Question 2014 6, Involving Organic Compounds

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Topic 7 Chemical Reactions Review 7 Dec 2017 Agenda FR Sample Question 2014 6, Involving Organic Compounds Topic 7 Chemical Reactions Review 7 Dec 2017 Agenda FR sample question 2014 6, involving organic compounds What’s in the bottle this time? Topic 7 Chemical Reactions Types of FR Qu 2. 2014 Atom Level Views Practice Test and Final Prep. Organic Related FR Question 2014? What is it really about? A student places a mixture of plastic beads consisting of polypropylene (PP) and polyvinyl chloride (PVC) in a 1.0 L beaker containing distilled water. After stirring the contents of the beaker vigorously, the student observes that the beads of one type of plastic sink to the bottom of the beaker and the beads of the other type of plastic float on the water. The chemical structures of PP and PVC are reprsented by the diagrams below which show segments of each polymer. a) Given that the spacing between polymer chains is similar, the beads that sink are made of which polymer? Explain. polyproplyene “Spacing similar” same a) Given lengththat the chains, spacing similar between polymer chains is similar,distance the beads apart that in 3 sink are made of which polymer?dimensions Explain. - this is same as saying if had an equal polyproplyene volume would have same number of polymer units of each Repeating unit CH CHCH vs a) Given that the2 spacing3 between polymer chains is CH CHCl similar, the beads2 that sink are made of which polymer? Explain.CH3 vs Cl polyproplyene 15 amu vs 17amu It’s a density question! The PVC beads sink because they are more dense than the PP beads. If the spacing between chains in similar in both polymers, similar volumes will contain similar repeating units of each polymer, so the fact that Cl has greater mass than CH3 ( 35.45 amu vs 15amu) must be sufficient to increase the density (mass/volume) of the PVC past that of water (1g/cm3). PP is synthesized from propene and PVC from vinyl chloride the structures of the molecules are shown below. vinyl chloride (chloroethene) b) The boiling point of liquid propene (225K) is lower than the boiling point of liquid vinyl chloride (260K). Account for the difference in terms of the types and strengths of intermolecular forces present in each liquid. vinyl chloride (chloroethene) All this for one point... Propene is essentially nonpolar with only London dispersion forces while vinyl chloride has both LDFs and dipole-dipole forces (as a result of the high electronegativity of the chlorine atom in the molecule). Vinyl chloride has a larger electron cloud, is more polarizable, and has a larger dipole moment. Thus, intermolecular attractions are stronger in vinyl chloride, which results in it having the higher boiling point. In a separate experiment, the student measure the enthalpies of combustion of propene adn vinly chloride. The student determines that the combustion of 2.00 mol of vinyl chloride releases 2300 kJ of energy, according to the equation below. 2 C2H3Cl(g) + 5O2(g) → 4 CO2(g) + 2 H2O + 2HCl c) Using the table of standard enthalpies of formation below, determine whether the combustion of 2.00 mol of propene releases more, less, or the same amount of energy that 2.00 mol of vinyl chloride releases. Justify your answer with a calculation. The balanced equation for the combustion of 2.00mol of propene is 2 C3H6(g) + 9 O2(g) → 6 CO2(g) + 6 H2O 2 C3H6(g) + 9 O2(g) → 6 CO2(g) + 6 H2O Using Hess’ Law o o o ΔH rxn = ΔH formation products - ΔH formation reactants o o o = 6(ΔH CO2) + 6(ΔH H2O) - 2(ΔH C3H6) o ΔH rxn = 6(- 394 kJ/mol) + 6(-242 kJ/mol) - 2 (21kJ/mol) = -3858 kJ/mol o Notice ΔH formation an element in its standard state is 0. The combustion of 2.00 mol. of propene releases more energy (-3900 kJ/mol) than the combustion of 2.00mol of vinyl chloride (-2300 kJ/mol). What’s in this bottle? Big Idea 3: Changes in matter involve the rearrangement and/or reorganization of atoms and/or electrons. Types of Chemical Reaction? Big Idea 3: Changes in matter involve the rearrangement and/or reorganization of atoms and/or electrons. Types of Chemical Reaction: Oxidation-reduction (redox) Precipitation Acid-base NaHCO3(s) + HC2H3O2(aq) → NaC2H3O2(aq) + H2O(l) + CO2(g) A student designs an experiment to study the reaction between sodium hydrogen carbonate and ethanoic acid. The reaction is represented by the equation above. The student places 2.24g of sodium hydrogen carbonate in a flask and adds 60.0 mL of 0.875 M ethanoic acid. The student observes the formation of bubbles and that the flask gets cooler as the reaction proceeds. a) Identify the reaction represented above as an acid-base reaction, precipitation reaction, or redox reaction. Justify your answer. (2014 Qu 2) NaHCO3(s) + HC2H3O2(aq) → NaC2H3O2(aq) + H2O(l) + CO2(g) a) Identify the reaction represented above as an acid-base reaction, precipitation reaction, or redox reaction. Justify your answer. According to the equation above there is no solid product so this cannot be a precipitation reaction. None of the oxidation numbers change so this is not a redox reaction. This is an acid-base reaction. (The weak acid HC2H3O2 reacts with the weak base HCO3- with HC2H3O2 donating a proton.) NaHCO3(s) + HC2H3O2(aq) → NaC2H3O2(aq) + H2O(l) + CO2(g) b) Based on the information above, identify the limiting reactant. Justify your answer with calculations. NaHCO3(s) + HC2H3O2(aq) → NaC2H3O2(aq) + H2O(l) + CO2(g) b) Based on the information above, identify the limiting reactant. Justify your answer with calculations. 2.24 g NaHCO3 NaHCO3(s) + HC2H3O2(aq) → NaC2H3O2(aq) + H2O(l) + CO2(g) b) Based on the information above, identify the limiting reactant. Justify your answer with calculations. 2.24 g NaHCO3 x 1mol NaHCO3 = 0.0267 mol NaHCO3 84.0g NaHCO3(s) + HC2H3O2(aq) → NaC2H3O2(aq) + H2O(l) + CO2(g) b) Based on the information above, identify the limiting reactant. Justify your answer with calculations. 60.0 ml x 0.875 M = 0.0525 mol HC2H3O2 1000 mL According to the equation the NaHCO3 and HC2H3O2 react in a 1:1 ratio, so the limiting reactant here is the NaHCO3. NaHCO3(s) + HC2H3O2(aq) → NaC2H3O2(aq) + H2O(l) + CO2(g) c) The student observes that the bubbling is rapid at the beginning of the reaction and gradually slows as the reaction continues. Explain this change in the reaction rate in terms of the collisions between reactant particles. NaHCO3(s) + HC2H3O2(aq) → NaC2H3O2(aq) + H2O(l) + CO2(g) c) As the reaction proceeds, both reactants are consumed and the number of particles remaining in the reaction vessel decreases (there is a smaller pile of sodium hydrogen carbonate and the concentration of the ethanoic acid decreases.) With fewer reactant particles available, collisions between them become less likely and the rate of reaction will decrease. NaHCO3(s) + HC2H3O2(aq) → NaC2H3O2(aq) + H2O(l) + CO2(g) d) In thermodynamic terms, a reaction can be driven by enthalpy, entropy, or both. i) Considering that the flask gets cooler as the reaction proceeds, what drives the chemical reaction. Answer by drawing a circle around one of the choices below. Enthalpy only Entropy only Both enthalpy and entropy NaHCO3(s) + HC2H3O2(aq) → NaC2H3O2(aq) + H2O(l) + CO2(g) d) In thermodynamic terms, a reaction can be driven by enthalpy, entropy, or both. i) Considering that the flask gets cooler as the reaction proceeds, what drives the chemical reaction. Answer by drawing a circle around one of the choices below. Enthalpy only Entropy only Both enthalpy and entropy NaHCO3(s) + HC2H3O2(aq) → NaC2H3O2(aq) + H2O(l) + CO2(g) d) ii) Justify your selection in part d(i) in terms of ΔGo ΔGo = ΔHo - TΔSo (on your reference sheet) Chemical reactions are thermodynamically favorable when ΔGo is negative. Since this reaction is endothermic (the flask gets cooler, ΔHo is positive), the reaction is not driven by enthalpy, because enthalpy does not help to make ΔGo negative. There are no gases in the reactions, but one of the products is a gas. This means that ΔSo will be positive, which is what helps make ΔGo negative and drives the reaction thermodynamically. e) The hydrogen carbonate ion has 3 carbon-oxygen bonds. Two of the carbon-to-oxygen bonds have the same length and the third carbon-to-oxygen bond is longer than the other two. The hydrogen atom is bonded to one of the oxygen atoms. In the box below, draw a Lewis electron-dot diagram (or diagrams) for hydrogen carbonate ions that is (are) consistent with the given information. e) don’t forget the lone (nonbonding) electron pairs f) A student prepares a solution containing equimolar amounts of ethanoic acid and sodium hydrogen carbonate. The pH of the solution is measured to be 4.7. The student adds two drops of 3.0M nitric acid and stirs the sample, observing the pH remains at 4.7. Write a balanced, net-ionic equation for the reaction between nitric acid and the chemical species that is responsible for the pH remaining at 4.7. Buffering. - + C2H3O2 + H3O → HC2H3O2+ H2O Atom level views Key Nitrogen atom Oxygen atom Atom level views - what do these represent? Atom level views - what do these represent? Oxygen molecule dinitrogen trioxide O2 N2O3 Use these models then to represent the reaction: Nitrogen gas reacts with oxygen gas to produce dinitrogen trioxide gas. Oxygen molecule dinitrogen trioxide O2 N2O3 Nitrogen gas + with oxygen gas →dinitrogen trioxide gas Nitrogen gas + with oxygen gas →dinitrogen trioxide gas Skeleton Equation N2 + O2 → N2O3 Nitrogen gas + with oxygen gas →dinitrogen trioxide gas Balanced Equation? N2 + O2 → N2O3 Nitrogen gas + with oxygen gas →dinitrogen trioxide gas Balanced Equation 2N2 (g) + 3O2(g) → 2N2O3 (g) Nitrogen gas + with oxygen gas →dinitrogen trioxide gas What does this represent? Nitrogen gas + with XS oxygen gas →dinitrogen trioxide Excess oxygen - still present in reaction vessel after the reaction stops.
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