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Course: CHEM 2B, Winter 2011
School: UC Davis
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Word Count: 1093

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Two Last Lectures First Law of Thermodynamics Heat Capacities allow calculations of energy transfer as heat Work is a mode of energy transfer Enthalpy is a state function that relates to the potential for heat transfer at constant pressure. Example: work depends on path We compare the work energy by expanding 1 mol of Ne from 1 to 2 atm in: a) One big expansion (2 to 1 atm) b) Two expansions (2 to 1.5 atm,...

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Two Last Lectures First Law of Thermodynamics Heat Capacities allow calculations of energy transfer as heat Work is a mode of energy transfer Enthalpy is a state function that relates to the potential for heat transfer at constant pressure. Example: work depends on path We compare the work energy by expanding 1 mol of Ne from 1 to 2 atm in: a) One big expansion (2 to 1 atm) b) Two expansions (2 to 1.5 atm, then 1.5 to 1 atm) Solution, step 1 First-the single-step expansion from 2 atm to 1 atm pressure. V= nRt , P 1 mol 298 K 0.0821 Vinitial = 2 atm 1 mol 298 K 0.0821 Vfinal = 1 atm L-atm mol K =12.2329 L; L-atm mol K =24.4658 L; work = -PV= 1 atm (24.4658 12.2329 L) = 12.2329 L-atm; or about -1235.5 Joules (1 L-atm 101 Joules) Solution, step 2 Now, the two-step expansion from 2 atm to 1.5 atm, then 1.5 to 1 atm pressure. V2 atm = 12.2329 L; V1atm =24.4658 L; (from the previous calculation) 1 mol 298 K 0.0821 V1.5 atm = 1.5 atm L-atm mol K =16.3105 L; work1 = -PV= -1.5 atm (16.3105 L- 12.2329) = -6.1164 L-atm work 2 = -1.0 atm (24.4658 16.3105 L) = -8.1553 L-atm; total work = -6.1164-8.1553 = -14.2717 L-atm; -1441.4 Joules We can create a state function from q and w because the internal energy, U, does not depend on path. U = q - PV H = U + PV = q at constant pressure H is a state function, meaning that it doesn't depend on path and can be tabulated. Enthalpy Heat released or absorbed at constant P Standard enthalpy 5 Pa on pure measured at 1 atm = 1.01325 x 10 substances symbol: H Coffee Cup Calorimeter A simple calorimeter. Well insulated and therefore isolated. Measure temperature change. qrxn = qcal = C (Tfinal Tinitial ) i.e. The heat of the reaction is simply the negative of the heat producing the T change in the calorimeter! Bomb Calorimeter qrxn = -qcal qcal = qbomb + qwater + qwires + qcal = CT How do we know C? Combustion of a standard: e.g. benzoic acid C6H5COOH Heat combustion = -3227 kJ/mol Rigid insulating walls fixed volume and isolated Measure T change heat of the chemical reaction Example, Heat of Reaction 0.1g of liquid benzene is burned in a closed calorimeter. The temperature of the calorimeter increases by 2.609 C. The calorimeter has a heat capacity of Cp=1.602 kJ/C) Find the molar heat of combustion of benzene. C6 H 6 (l ) + 15 2 O 2 (g ) . 6CO 2 (g ) + 3H 2 O(l ) Solution Heat Released: qrxn = qcal qcal = C T = 1.602 103 J / C 2.609C = 4179.618 J qrxn = 4179.618 J Please note that it is exothermic! Molar heat of combustion: # moles C6 H 6 burned, n = qrxn qrxn 4179.618 J = = 0.00128mol n 0.1g 0.1g = = 0.00128mo (6 12.011+6 1.0079) g / mol 78.1134 g / mol = - 3.265106 J / mol = 3265 kJ / mol Phase changes are a simple reaction Energy is needed to break bonds: solid U liquid U gas Total q can be calculated from Cp, Csp, Hvap and Hfus temperatures are constant during phase changes Example Calculate the total heat needed to change 18 g (1 mol) of ice at -25 C to vapor at +125 C. Csp(Ice) = 2.09 J/gC Hfus = 6.01 kJ/mol Csp(Water) = 4.184 J/gC Csp(Vapor) = 1.84 J/gC Hvap = 40.7 kJ/mol o Step 1--heating of ice from -25 to 0 C J = 2.09 J/gC x (0- (-25))oC 18 x g of Ice = 940 J o Step 2-- melting ice at 0 C J = Hfus = 6.01 kJ/mol x 1 mol H2O = 6010 J Step 3- heating water from 0 C to 100oC o J = 4.184 J/gC x 18 g x (100-0) C = 7531.2 Joules o Step 4 - boiling water at 0 C = 40.7 kJ/mol x 1 mol = 40,700 Joules Step 5- heating water vapor from 100 C to 125oC o J = 1.84 J/gC x 18 g x (125-100)oC = 828 Joules Add the heats for each step: total heat = 940 J + 6010 J + 7531 J + 40,700 J + 828 J Each element has a reference form that is defined to have enthalpy of formation equal to zero if at standard conditions. Hf (O2(g)) 0 kJ/mol Hf (H2(g)) 0 kJ/mol This definition allows chemists to build a table of enthalpies from calorimetrically measured heats and from heat capacities. Standard Enthalpy of Formation, Hf The enthalpy change that occurs in the formation of one mole of a substance in the standard state from the reference forms of the elements in their standard states. Standard state = 1 bar pressure H2(g) + O2(g) H2O(l) Hf (H2O(l)) = -285.83 kJ/mol Reference Forms of the Elements O: O2(g) - a diatomic gas at room temperature and 1 bar 3O2 Br: Br2(l) - a diatomic liquid at room temperature and 1 bar Br2(l) C: Br2(g), H=+30.9 kJ C(graphite) C(graphite) S: H=+285 kJ 2O3, C(diamond), H=+1.9 kJ S8(s) - many other allotropes, e.g. S7 Usually the reference state is the most stable form. Exception: P: P4(s,white) P4(s,white) P(s,red), H=-17.6 kJ Hess Law (this is powerful) Reaction enthalpies are additive: Hf(reverse) = -Hf(forward) Hcondensation = Hliquid - Hvapor You can use them in a balanced chemical reaction. Hesss Law and balanced chemical reactions N2(g) + O2(g) NO(g) H = +90.25 kJ NO(g) + O2(g) NO2(g) H = -57.07 kJ N2(g) + O2(g) NO2(g) H = +33.18 kJ Tabulated enthalpies can be used among different reactions. This tabulation lets you predict things, like heats of reactions and even the temperature dependencies of equilibria. Hess Law If a process occurs in stages or steps (even hypothetically), the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps. A ? AC B H (AC) H (DB) C D H (CD) H (AC) CD H (CD) DB H (DB) AB H (AB) Recall: H is a state function: path independent! Reaction Enthalpy Reactants Products Hf =nHf (products) - nHf (reactants) Hf positive Endothermic Hf negative Exothermic Hf Standard enthalpy of formation H is extensive: H = U + PV PCl5(l) PCl3(l) + Cl2(g); Ho = +123.8 kJ Can multiply: 4PCl5(l) 4PCl3(l) + 4Cl2(g); Ho = 4(+123.8 kJ) = +495.2 kJ Can reverse: PCl3 (l) + Cl2 (g) PCl5 (l); H = -123.8 kJ Example - Combustion Is the combustion of propane C3H8(g) endothermic or exothermic? Hf (O2) = 0 kJ/mol Hf (H2O) = -285.83 kJ/mol Hf (CO2) = -393.51 kJ/mol Hf (C3H8) = -2220 kJ/mol Solution Reaction: Complete combustion all propane is converted to carbon dioxide (gas) and water (liquid at room temperature) C3 H 8 ( g ) + 5O2 ( g ) 3CO2 ( g ) + 4 H 2O (l ) Enthalpy : H combustion = 3H f (CO2( g ) ) + 4H f ( H 2O( l ) ) H f (C3 H 8( g ) ) 5H f (O2( g ) ) = (-1180.53) + (-1143.32) (-103.8) = -2220 kJ/mol, EXOTHERMIC (of course) (0) kJ/mol
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