Energy and Chemical Reactions - Lecture Slides | CHEM 1307, Study notes of Chemistry

Download Energy and Chemical Reactions - Lecture Slides | CHEM 1307 and more Study notes Chemistry in PDF only on Docsity! Energy & Chemical Reactions Energy Most of the energy we use in our daily lives is obtained by carrying out chemical reactions (ie burning fossil fuels). Thermodynamics is the science of heat and work. Energy is defined as the capacity to do work. You do work against the force of gravity when carrying yourself and hiking equipment up a mountain. Food is a source of chemical energy – when we eat, the energy stored in the compounds is released as the food is metabolized. System and Surroundings An object or collection of objects being studied. System: Everything outside of the system. Surroundings: the surroundings the system Direction of Energy Transfer When energy is transferred as heat between a system and its surroundings, the direction is described as endo or exothermic. Energy is transferred from the system to its surroundings. E of the system decreases, E of the surroundings increases. Exothermic: Energy is transferred from the surroundings to the system. E of the system increases, E of the surroundings decreases. Endothermic: Energy Units Nearly all chemical changes involve an energy transfer, typically in the form of heat. The SI unit for energy is the joule. 1 J = 1 kg m2 s2 If you drop a 6-pack of soda on your foot, the KE at the moment of impact is between 4-10 J. 4.184 J = 1 cal A calorie is defined as the amount of energy required to raise the temperature of one gram of water by 1 °C. Joules and calories are small quantities, so we use kJ and kcal. Calories from food energy are actually kcal. Specific Heat Capacity Implications When bread is wrapped in aluminum foil and heated in the oven, you can remove the foil with your fingers. The bread and foil are very hot, but only a small mass of Al foil is used and it has a low specific heat capacity, so only a small amount of energy is transferred to your fingers. How much energy must be transferred to raise the T of a cup of coffee (250 mL) from 20.5 °C to 95.6 °C? Use the specific heat capacity for water. q = C x m x DT q = (4.184 J/gK) x (250 g) x (368.6 K – 293.5 K)= 78554 J = 79 kJ Positive value means E transferred to the coffee as heat. Specific Heat Capacity Practice We can use energy transfer as a way to determine the specific heat capacity of a given unknown substance. A 55.0 g piece of metal is heated in boiling water to 99.8 °C then dropped into a cold water bath containing 225 g of water with an initial temperature of 21.0 °C. The final temperature of both substances is 23.1 °C. What is the C of the metal? Specific Heat Capacity Practice Things to Think About: 1. Metal and water are the system, beaker and lab are the surroundings. 2. The metal and water end up at the same temperature. 3. Assume energy is transferred only as heat. 4. E transferred from the metal to the water (qmetal) is negative due to the change in temp; qwater is positive because of the increase in temp. 5. qwater + qmetal = 0 or -qmetal = qwater (equal but opposite sign) [(4.184 J/gK)(225 g)(296.3 K – 294.2 K)] + [(C)(55.0 g)(296.3 K – 373.0 K)] = 0 Cmetal = 0.469 J/gK Energy and State Changes Temperature is constant throughout a change of state. During a change of state, the added E is used to overcome molecular forces, not to increase the temperature. Note that T is constant as ice melts Energy and State Changes What is the energy required to convert 500 g of ice at -50 °C to steam at 200 °C? Heat of Fusion = 333 J/g Heat of Vaporization = 2256 J/g Step 1: Warm ice from -50 °C to 0 °C. q = C x m x DT = (2.06 J/gK)(500 g)(273 K – 223 K) = 5.15 x 104 J Step 2: Melt ice at 0 °C. q = m x Hf = (500 g)(333 J/g) = 1.67 x 10 5 J Step 3: Warm liquid from 0 °C to 100 °C. q = C x m x DT = (4.184 J/gK)(500 g)(373 K – 273 K) = 2.09 x 105 J Step 4: Boil water at 100 °C. q = m x Hv = (500 g)(2256 J/g) = 1.13 x 10 6 J Step 5: Heat water vapor from 100 °C to 200 °C. q = C x m x DT = (1.86 J/gK)(500 g)(473 K – 373 K) = 9.3 x 104 J qtotal = q1 + q2 + q3 + q4 + q5 = 1.65 x 10 6 J Energy and State Changes What is the minimum amount of ice at 0 °C that must be added to a can of diet soda (340 mL) to cool the soda from 20.5 °C to 0 °C? Use the specific heat capacity and density of water. qcola + qice = 0 Ccola x m x DT + qice = 0 [(4.184 J/gK)(340 g)(273 K – 293.5 K) + [(333 J/g)(mice)] = 0 333 x mice = 29162.5 mice = 87.6 g If more ice is used, the final temperature will still be 0 °C when thermal equilibrium is reached, but some ice will remain. If less ice is used, the final temperature will be greater than 0 °C and all the ice will be melted. OWL Problem A 31.80 g sample of tin is initially at 221 °C. If 5014 J of heat are added to the sample at constant pressure, which of the following are true? DHfus (232 °C) = 59.6 J/g DHvap (2270 °C) = 1939 J/g Csolid = 0.226 J/g°C Cliquid = 0.243 J/g°C a. The sample is at a temp greater than 232 °C. b. The sample is at exactly 232 °C. c. The sample is a solid in equilibrium with a liquid. d. The sample is a gas. e. The sample is a liquid. q1 = 79 J, q2 = 1895 J, q3 = 15748 J a & e First Law of Thermodynamics The energy change for a system (DU) is the sum of the energy transferred as heat (q) and the energy transferred as work (w). Work: Energy transfer that occurs as a mass is moved against an opposing force. If a system does work on its surroundings, Esystem will decrease. If work is done by the surroundings, Esystem will increase. internal energy change work energy transferred DU = q + w heat energy transferred Internal Energy Changes Change Sign Convention Effect of Usystem E transferred as heat to system (endothermic) q > 0 (+) U increases E transferred as heat from the system (exothermic) q < 0 (-) U decreases E transferred as work done on system w > 0 (+) U increases E transferred as work done by system w < 0 (-) U decreases P-V Work: Work associated with a change in volume (DV) that occurs against a resisting external pressure (P). When P is constant: w = -P x DV Internal Energy Problems An automobile engine provides 620 J of work to push the pistons. In the process, the internal energy changes by -2960 J. Calculate the amount of heat carried away by the cooling system. DU = q + w q = DU - w Work is done by the system so w (-) q = -2960 J – (-620 J) = -2340 J q < 0 so heat is given off by the system Enthalpy Heat energy transferred at constant pressure. DU = qp + wp DU = qp - PDV qp = DU + PDV DH = DU + PDV DH = qp DV is typically very small therefore work is small. DH and DU differ by PDV. DV is large when gases are formed or consumed. DH (-) E transfer from the system, DH (+) E transfer to the system. State Functions A state function depends on the initial and final state of the system but is independent of the path taken. (Doesn’t matter how you got to the end product!) No matter how you go from reactants to products in a reaction, the values of DH and DU are always the same. P, V and T are also state functions q and w are not state functions Enthalpy How many grams of H2O2 (l) would have to react to produce 24.5 kJ of energy? 2 H2O2 (l) 2 H2O (l) + O2 (g) DrH = - 196 kJ/mol 24.5 kJ x 2 mol H2O2 196 kJ = 8.5 g H2O2x 34 g H2O2 1 mol H2O2 Ethene and hydrogen react to form ethane. How much heat is released when 15.6 g of ethane form? C2H4 (g) + H2 (g) C2H6 (g) DrH = - 137 kJ/mol 15.6 g C2H6 x 1 mol C2H6 30 g C2H6 = 71 kJx 1 mol C2H4 1 mol C2H6 x 137 kJ 1 mol C2H4 Constant Pressure Calorimetry The experimental determination of the enthalpy change of a reaction. Exothermic: Twater increases Endothermic: Twater decreases qreaction + qwater = 0 Assume no E transfer beyond solution. Coffee Cup Calorimetry A chunk of lead (C = 0.159 J/g°C) weighing 18.8 grams and originally at 97.64 °C is dropped into an insulated cup containing 78.6 grams of water at 22.86 °C. What is the final temperature of the water? qwater + qlead = 0 [Cw x mw x DT] + [CPb x mPb x DT] = 0 [4.184 J/gC x 78.6 g x (Tf – 22.86)] + [0.159 J/gC x 18.8 g x (Tf – 97.64)] = 0 [328.9Tf – 7517.8] + [2.99Tf – 291.9] = 0 331.9Tf = 7809.7 T = 23.5 °C Hess’s Law DH for an overall process is the sum of the DH values of all the individual steps. C(graphite) + ½ O2 (g) CO (g) Some reactions are complicated and DH cannot be measured C(graphite) + ½ O2 (g) CO (g) DH1 = ? We can combine multiple reactions to calculate DH CO (g) + ½ O2 (g) CO2 (g) DH2 = -283 kJ/mol C(graphite) + O2 (g) CO2 (g) DH3 = -393 kJ/mol DH3 = DH1 + DH2 DH1 = DH3 - DH2 DH1 = -393 kJ/mol – (-283 kJ/mol) = -110 kJ/mol Hess’s Law Calculate the DH for the formation of methane from graphite and hydrogen gas. Desired reaction: C(s) + 2 H2 (g) CH4 (g) DHrxn = DH1 + DH2 + DH3 Eqn 1: C (s) + O2 (g) CO2 (g) DH1 = -393 kJ/mol Eqn 2: H2 (g) + ½ O2 (g) H2O (l) DH2 = -286 kJ/mol Eqn 3: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) DH3 = -890 kJ/mol C (s) + O2 (g) CO2 (g) DH1 = -393 kJ/mol 2 H2 (g) + O2 (g) 2 H2O (l) 2 x DH2 = -572 kJ/mol CO2 (g) + 2 H2O (l) CH4 (g) + 2 O2 (g) DH3 = +890 kJ/mol C(s) + 2 H2 (g) CH4 (g) DHrxn = -393 kJ/mol + -572 kJ/mol + 890 kJ/mol = -75 kJ/mol Standard Molar Enthalpies of Formation The enthalpy change for the formation of 1 mol of a compound directly from its elements in their standard state. For an element, DHf = 0 If DHf is negative, formation of the compound is exothermic is positive, formation of the compound is endothermic The more negative the DHf, the more stable the compound Compound DHf (kJ/mol) HF (g) -273.3 HCl (g) -92.31 HBr (g) -35.29 HI (g) +25.36 Which of the hydrogen halides is most stable? DHf does not always describe a reaction that can be done in the lab Enthalpy Change for a Reaction DHrxn = S DHf (products) – S DHf (reactants) Add up all the DHf for the products multiplied by their stoichiometric coefficients and subtract from the sum of the products. How much heat is required to decompose 1 mol of CaCO3? Compound DHf (kJ/mol) CaCO3 (s) -1207.6 CaO (s) -635.1 CO2 (g) -393.5 CaCO3 (s) CaO (s) + CO2 (g) DHrxn = [-635.1 kJ/mol + -393.5 kJ/mol] – [-1207.6 kJ/mol] DHrxn = + 179 kJ/mol (decomposition of CaCO3 endothermic!) Enthalpy Change for a Reaction Calculate the enthalpy change that occurs when 10.0 g of nitroglycerin are detonated. DHf[nitroglycerin] = -364 kJ/mol From Appendix L: DHf[CO2(g)] = -393 kJ/mol DHf[H2O(g)] = -242 kJ/mol DHf[N2(g)] = 0 kJ/mol DHf[O2(g)] = 0 kJ/mol 2 C3H5(NO3)3 (l) 3 N2 (g) + ½ O2 (g) + 6 CO2 (g) + 5 H2O (g) Solve for DHrxn: DHrxn = S DHf (products) – S DHf (reactants) DHrxn = [3 mol N2 x 0 kJ/mol + ½ mol O2 x 0 kJ/mol + 6 mol CO2 x -393 kJ/mol + 5 mol H2O x -242 kJ/mol] – [2 mol C2H5(NO3)3 x -364 kJ/mol] = -2840 kJ/mol rxn Calculate moles of nitroglycerin: 10.0 g C3H5(NO3)3 x 1 mol C3H5(NO3)3 227.1 g C3H5(NO3)3 = 0.0440 mol C3H5(NO3)3 Calculate enthalpy change for 0.0440 mol C3H5(NO3)3: 0.0440 mol ng x 1 mol rxn 2 mol ng = -62.5 kJx -2840 kJ 1 mol rxn