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Lecture Notes 2 DELAAT copy.pdf

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University of Guelph
CHEM 1050
Richard Delaat

– PART 2 – NOTES: Thermochemical Equations A chemical equation may look like the following: CH + 2O4 CO +22H O 2 2 A THERMOCHEMICAL equation looks like the following: CH 4(g) + 2O 2(g)  CO 2(g) + 2H O2 (l, H = –890kJ What is different?  States are ____________ specified: (ssolid, (lliquid, (g)gas,(aq)aqueous.  H values (usually kJ mol ) are signed (+/–).  H is a _______________________ term. This thermochemical equation tells us that when 1 mole of methane is burned completely in air to give CO g2s and liquid water, 890 kJ of energy is given off. Convention: Measurements usually made at 25C (298K). A typical ENDOTHERMIC reaction: CaCO 3(s) CaO (s)+ CO 2(g, H = +178 kJ i.e., we must supply 178 kJ of energy to convert 1 mole CaCO (s) 3 to CaO (s)and CO 2(g. Some Simple Thermochemical Rules & Laws RULE 1: Thermochemical equations normally tell you about energy changes associated with _______________ of something. e.g., We saw above that the H for combustion of 1 mole CH 4 at 298 K was –890 kJ. When a thermochemical equation is multiplied by any factor, the value of H is ____________________________. Question: How much energy is released if we burn 1.00 L CH (g)at 5 4 1.01×10 Pa and 25C? The first thing we should ask ourselv5s is "How many moles of CH 4s present in 1.00 L at 1.01×10 Pa & 25C?" Ideal Gas Law: Note: If P in Pa and V is m , then use R = 8.314 Pa∙m ∙K mol –1 –1 If P in kPa and V is L, then us R = 8.314 kPa∙L∙K mol–1 –1 15 5 # moles of CH 4g)at 1.01×10 Pa and 25C: NOTES: Since PV = nRT, therefore n = PV/RT n = Based on the thermochemical equation given earlier, 1 mole CH 4 evolves __________ kJ, therefore 4.08×10 –2 moles of CH e4olves –2 4.08×10 mol × __________ kJ/mol = 36.3 kJ RULE 2: If we REVERSE the direction of a reaction, we _________________ of H e.g. CaCO 3(s)  CaO (s) + CO 2(g, H = +178 kJ CaO (s) + CO 2(g)  CaCO 3(s, H = _______________ kJ RULE 3: Thermochemistry (and Thermodynamics) is only interested in energy differences between initial and final states. It is not concerned with HOW (i.e., "by what mechanism") a reaction takes place. All this means is that we can add (or subtract) thermochemical equations to produce other equations. Hess's Law: Thermochemical equations can be added (or subtracted) to yield other thermochemical equations. (more on this later) THERMOCHEMISTRY OF CHANGE 1. Changes of Physical State e.g., solid  liquid  gas OR solid  gas A. Enthalpy of Fusion, H :fuse enthalpy change when we fuse (i.e., melt) 1 mole of something. i.e., 1 mole solid  e.g., H fus is always ______________________; energy had to be added to weaken forces (_______________) and cause the solid to flow. Note: T remains constant while process occurring, e.g., 0C for 2 O(s. 16 The "" symbol refers to “STANDARD  CONDITONS”: NOTES: 5  pure reactants at 1 atm of pressure (actually 1 bar which equals 10 Pa)  pure products at _____________________________.  solids & liquids are in their most stable form, known as "standard state"  solutes are ________________________.  temperature must be specified. Note: The most common temperature used is 25C BUT it is not included within the definition of "standard" conditions. For the reverse process: H2O (l) H O 2 (s; H = 0C 0C What is the name for this process? _________________________________ i.e.In general, H = –H Reverse process Forward process B. Enthalpy of Vaporization, H vapenthalpy change when we vapourise 1 mole of liquid to gas. (H is always _________________) vap e.g., H2O (l) H O 2 (g); Hvap= 100C 100C What is the name for the reverse process? Heating Curve for Water gas 100C boiling H vap point Temp (C) liquid 0C H fus melting solid point Time heat supplied 17 C. `Enthalpy of Sublimation, H sub: NOTES: – enthalpy change when we sublime 1 mole solid  1 mole _____________ (H subis always ______________________) e.g., Note: H sub= H fus+ Hvap ...can do this since enthalpy is a state function. 2. Enthalpy Changes for Chemical Reactions A. Enthalpy of Combustion, H comb – enthalpy change when we burn ____________________ completely in O . 2 e.g., C(s)+ O 2(g) CO 2(g; H comb = – 393.5 kJ H combis always ___________________, because energy is always _____________________ (exothermic). B. Enthalpy of atomization: – enthalpy change associated with the formation of _____________________ from its elements. e.g., ½H 2(g)  H (g; H = +218 kJ Note: Enthalpy of reaction or "Heat" of reaction is a general description used to refer to any reaction. HESS' LAW (Refer to page 241 in text.) Hess' Law – H for an overall process is the sum of the H's for the individual steps of the process. Thermochemistry (and thermodynamics) is only interested in the energy differences between initial and final states. It doesn't matter how we got there. Hence we can added (or subtracted) thermochemical equations to produce other thermochemical equations. (Thermochemical Rule # _____) Example: Suppose we wanted to know H for Sn(s)+ 2Cl (g)  SnCl (l, but you could Rxn 2 4 not find the value in any set of tables. But, you did find the following: (a) Sn(s)+ Cl 2(g)  SnCl 2(s; HRxn = –350 kJ (b) SnCl2(s) + Cl 2g)  SnCl 4(l; H Rxn = –195 kJ Can you manipulate these equations to arrive at the one we want? 18 NOTES: Sn(s) + 2Cl 2(g)  SnCl 4(,) H Rxn= __________ kJ Example: Given the following data: 3 N 2(g)+ 2O 2(g)  N O 2 3(g) H = 83.7 kJ N 2(g)+ O 2(g) 2NO (g) H = 180.4 kJ ½N 2(g)+ O 2g)  NO 2(g) H = 33.2 kJ what is H for the rxn: N O (g)  NO (g)+ NO (g) 2 3 2 A) –207.1 kJ D) 39.7 kJ B) –39.7 kJ E) 207.1 kJ C) 24.3 kJ Extra Problem: Given: (a) ½ N2(g)+ ½ O 2(g)  NO (g; H Rxn = +90.4 kJ (b) ½ N2(g) + O 2(g)  NO 2g); HRxn = +33.2 kJ Calculate H Rxn for: 2NO(g) + O 2g)  2NO (g2 Ans: H Rxn= – 114.4 kJ To use Hess's Law effectively we obviously need to have access to tables of H values. Rather than write out full thermochemical equation for every H, we devise a method (outlined below) which will allow you to deduce the appropriate thermochemical equation. STANDARD ENTHALPY OF FORMATION (H ) f The Standard Molar Enthalpy of Formation (or the standard heat of formation) is the enthalpy change when we form ONE mole of a substance from its ELEMENTS in their STANDARD STATES. Aside: Previous, we looked at the combustion reaction: C (s)+ O 2(g) CO 2g); Hcomb = – 393.5 kJ This reaction also leads to the formation of CO(g)in its standard state. 2 Hence, the heat of this reaction (–393.5 kJ/mol), which we noted as the heat of combustion, is also the heat of formation of C2(g, i.e., H fCO 2(g] = –393.5 kJ/mol 19 These H vflues can be obtained for many, many compounds (see Appendix C: NOTES: Thermodynamic Quantities for Substances and Ions at 25C). H  caf have either negative or positive values, e.g., Different forms of bromine H (kJ/mol) f Br2(l) 0 Br2(g) 30.91 Br(g) 111.88 Note: The energy difference between B2 (g) and Br(g) corresponds to the energy to break the Br-Br bond. Allotropes of Carbon H (fJ/mol) C (s, graphite) 0 C(s, diamond) +1.897 You should be able to write an equation that defines a heat of formation. Example: The heat of formation (H
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