Dr. C. Yau Spring 2015

Energy Part III:Calculation of ΔH froma) Thermochemical Equationsb) Heat of FormationChapter 6 Sec 6 – Sec 8of Jespersen 7TH ed Dr. C. Yau Spring 2015 1

Standard Heat of Reaction ΔH = “enthalpy of reaction” or “heat of reaction” = heat transferred in a rxn (usually in kJ, not kJ/mol) ΔHo= standard heat of reaction = ΔH at standard conditions o means standard conditions (1 atm, 1M if aq soln, usually 25oC) Remember these conditions! 2

Thermochemical Equations N2(g) + 3H2(g) 2NH3 (g) ΔHo = -92.39 kJ This tells us that 1 mole of N2 would produce 92.39 kJ of heat, that 3 moles of H2 would produce 92.39 kJ of heat, that production of 2 moles of NH3 would be accompanied by a release of 92.39 kJ of heat. NOTE that ΔHo is in kJ and not kJ/mol. The amt of heat transferred is directly proportional to the # moles in thermochemical eqn shown. 3

Thermochemical Equations Example: Magnesium burns in air to produce a bright light and is often used in fireworks displays. 2 Mg (s) + O2(g) 2MgO (s)Ho= -1203 kJ How many grams of Mg is needed to produce 400. kJ of heat? p.295 #6.75 How much heat (in kJ) is liberated by the combustion of 6.54 g of Mg? Set these problems up in dimensional analysis.

Thermochemical Equations N2(g) + 3H2(g) 2NH3 (g) ΔHo = -92.39 kJ The value of ΔHo depends on the coefficients in the equation. If coefficients are doubled, ΔHo would be doubled: 2N2(g) + 6H2(g)4NH3 (g) ΔHo = -92.39x2 kJ Note also that ΔHo is dependent on the physical states as stated in the equation. CH4(g) + 2O2(g) CO2(g) + 2H2O(l)ΔHo = - 890.5kJ CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔHo = - 802.3kJ 5

Example 6.6 p.276 The following thermochemical equation is for the exothermic reaction of hydrogen and oxygen that produces water. 2H2(g) + O2(g) 2H2O (l)ΔHo = -571.8kJ What is the thermochemical equation for this rxn when it is conducted to produce 1.000 mol H2O? Do Pract Exer 15 & 16 p.277

Thermochemical Equations If we reverse a reaction, the magnitude of ΔHo is the same but the sign is changed: C (s) + O2(g) CO2(g)H° =  393.5 kJ CO2(g) C (s) + O2(g)H° = + 393.5 kJ

Determination of ΔH by manipulation of Eqns. Use the two equations below to determine the standard enthalpy change for the reaction: H2O2(l) H2O (l) + ½ O2(g) (1) H2(g) + O2(g) H2O2(l)H° = 188 kJ (2) H2(g) + ½ O2(g) H2O (l)H° = 286 kJ Strategy: We need H2O2 on the left side, so Eqn 1 must be reversed. Eqn 2 probably can stay as is. WE MUST CHECK. How? On my exams you are expected to “show your work” as we are doing here in class. Take notes!

Determination of ΔH by manipulation of Eqns. Ethylene glycol, HOCH2CH2OH, is used as antifreeze. It is produced from ethylene oxide, C2H4O, by the reaction (1) C2H4O (g) + H2O (l) HOCH2CH2OH (l) What is the heat of reaction of this reaction... Given: (2) 2C2H4O (g) + 5O2(g) 4CO2(g) + 4H2O (l) H°= 2612.2 kJ (3) HOCH2CH2OH(l) + 5/2 O2(g)2CO2(g) +3H2O(l) H°= 1189.8 Kj What is the strategy? Do Pract Exer 19 & 20 p.282

Determination of ΔH by manipulation of Eqns. 2Cu (s) + O2(g) 2CuO (s)H° = 310 kJ 2Cu (s) + ½ O2 (g) Cu2O (s)H° = 169 kJ Use the two equations above to determine the H° of this reaction: Cu2O (s) + ½ O2(g) 2CuO (s) Is this exothermic or endothermic? Solve the problem by manipulating the given eqns. ANS -141 kJ

Determination of ΔH by manipulation of Eqns. Example 6.8 p.281 Fe2O3(s) + 3CO (g) 2Fe(s) + 3CO2(g)H° =  26.7 kJ CO(g) + ½ O2(g) CO2(g)H° =  283.0 kJ Calculate the value of H° for the following reaction: 2 Fe (s) + O2(g) Fe2O3(s)

Enthalpy Diagrams C (s) + ½ O2 (g) CO (g)H° = 110.5 kJ Construct an enthalpy diagram for the reaction. Learn the terminology, "enthalpy diagram." Know what is asked for on an exam.

Enthalpy Diagrams N2(g) + O2(g) 2 NO (g)H° = +181 kJ Draw the enthalpy diagram for this reaction. We are skipping Example 6.7, Pract Exer 17 & 18. You do not need to know how to draw enthalpy diagrams of that sort. However, you should know how to add equations together to determine the enthlapy change for the overall reaction. You should know how H diagrams differ between an endo- and exothermic reactions.

Hess' Law The value of H° for any reaction that can be written in steps equals the sum of the values of H° of each of the individual steps. This is based on the fact that enthalpy is a state function. The implication is that regardless of how many steps are taken the overall enthalpy change is the same.

Enthalpy as a State Function Enthalpy (H) depends only on its current state and not the path taken to get there. It is not affected by how many steps are used to get there. A  B C F D E H1 H2 H6 H3 H5 H4 H1 = H2 +H3 +H4+H5 +H6

Hess’ Law is extremely useful because if H1 cannot be measured, it can be calculated from H2 and H3. For example: A + B C D + E H1=? H3 H2 H1 = H2 + H3

Hess' Law Watch out for the directions of the arrows. What is H1 in terms of H2 and H3? F + G J K + L H1 H2 H3 H1 = H2-H3

Standard Heat of Combustion ΔHoc = standard heat of combustion It is the amount of heat released when one mole of a fuel substance is completely burned in pure oxygen gas with all reactants and products brought to 25oC and 1 bar pressure (1 atm). Combustion reactions are always exothermic. Therefore, ΔHoc is always negative.

Example 6.9 p.283 How many moles of carbon dioxide gas are produced by a gas-fired power plant for every 1.00 MJ (megajoule) of energy it produces? The plant burns methane, CH4 (g), for which ΔHoc is -890 kJ/mol Do Pract Exer 21 & 22 p.284

Standard Enthalpy of Formation LEARN THIS DEFINITION! ΔHfo= standard enthalpy of formation It is the amount of heat absorbed or evolved when specifically one mole of substance is formed at 25oC at 1 atm from its elements in their standard states.

ΔHfofor solid potassium sulfate is -1433.7 kJ. Write the thermochemical equation corresponding to this value. ΔHfofor solid ammonium chloride is – 315.4 kJ. Write the thermochemical equation corresponding to this value. ΔHfofor solid calcium hydroxide is -986.6 kJ. Write the thermochemical equation corresponding to this value. Do Example 6.10 and Pract Exer 23 & 24 p.286

ΔHfo of Elements Write the thermochemical equation corresponding to the ΔHfoof chlorine gas. What do you think the value of ΔHfowould be for chlorine gas? What about the value of ΔHfoof solid silver? of liquid mercury? Remember this! You will not be provided ΔHfoelements in their standard states. They are always ZERO kJ.

Applying Hess' Law to Heats of Formation Hrxn = ? A + B C + D Hf (D) Hf (C) Hf (A) Hf (B) All elements in their standard states. What is heat of reaction of A+B C+D? Hrxn = -Hf (A) -Hf (B) + Hf (C)+Hf (D) Hrxn = - {Hf (A) +Hf (B) } + {Hf (C)+Hf (D)} Hrxn = +{Hf (C)+Hf (D)} - {Hf (A) +Hf (B) } Hrxn = SumH (products) - Sum H(reactants)

Hess' Law of Summation ΔHfo values are given in Table 7.2 p. 285 and will be provided on exams. Remember Hf is per mole. You must multiply it with n (the number of moles from the coefficients of the balanced equation).

Example 6.11 p.287 Some chefs keep baking soda, NaHCO3, handy to put out grease fires. When thrown on the fire, baking soda partly smothers the fire and the heat decomposes it to give CO2, which further smothers the flame. The eqn is 2NaHCO3(s) Na2CO3(s) + H2O (l) + CO2(g) Use the data in Table 7.2 (p.285) to calc the ΔHo for this reaction in kilojoules. Do not put out a kitchen fire with water! http://www.youtube.com/watch?v=sZGzbd0IvUE&feature=related

Calculating ΔH For Reactions Using ΔHf° 2Fe(s) + 6H2O(l)→ 2Fe(OH)3(s) + 3H2(g) ΔH°f = -285.8 -696.5 in kJ mol-1 CO2(g) + 2H2O(l)→ 2O2(g) + CH4(g) ΔH°f =-393.5 -285.8 -74.8 in kJ mol-1 Do Pract Exer 25, 26 & 27 p. 288

SUMMARY What are the different ways you can determine the ΔH of a reaction? • Measure q from calorimetry expt.In open containers, q = ΔH • Calculate by manipulating given thermochemical equations. • Calculate from ΔHfo • Calculate from bond energies (handout) Also, remember how to calculate amt of heat from stoichiometry & thermochemical eqn.

Dr. C. Yau Spring 2015