Chem. 412 – Phys. Chem. I

Chem. 412 Phys. Chem. I

Sign Convention

System

Surroundings U = Internal Energy. q = heat flow; transfer of energy between two objects. w = work; product of force applied to an object over a distance.

Work Basic FormulationWork Done = Force x (Distance Moved)Differentially => dw = F dS

Work Gas Expansions and Compressions I (F14)

Work Gas Expansions and Compressions II (F14)

Work Gas Expansions and Compressions III (F14)

Work Gas Expansions and Compressions IV (F14)

Work Gas Expansions and Compressions V (F14)

Work Gas Expansions and Compressions I (F13)Work is area under curve of P-V Diagram. dw = F dS = - P dV

Work Gas Expansions and Compressions II (F13)Work is area under curve of P-V Diagram. dw = F dS = - P dV

Work Gas Expansions and Compressions III (F13)Work is area under curve of P-V Diagram.Reversible versus Irreversible Work

Work Gas Expansions and Compressions IV (F13)Work is area under curve of P-V Diagram.Reversible versus Irreversible Work

Work Gas Expansions and Compressions V (F13)Reversible versus Irreversible Work

Tutorial Problem on WorkConsider one mole of an ideal gas kept in a right-circular cylinder with a movable piston cap at constant temperature. The cap is attached to an external engine that is capable of moving the piston in either directions.Initially, the cylinder has a volume of 1.0 m3 at a pressure of 10. Pa. The cap was moved to give a final pressure of 1.0 Pa. In this case, the piston cap was moved at an infinitely slow rate to achieve this final pressure, thereby following the ideal gas law. Calculate the work involved in this process. This work is referred to as reversible work, wrev . [ 23 J ]The initial and final volume/pressure was kept the same as in part (a) but the piston cap was suddenly released to get to the final pressure value. Calculate the work involved in this process. This work is referred to as irreversible work, wirrev . [ 9 J ]Draw one P-V diagram for parts (a) and (b).Repeat parts (a), (b), and (c) by reversing the initial and final volume/pressure conditions; that is, compression instead of expansion. [ 23 J, 90 J ]Compare/Contrast/Discuss the above results.

First Law of Thermodynamics

First Law of Thermodynamics U = Internal Energy = Heat Flow under Constant Volume H = Enthalpy = Heat Flow under Constant Pressure H = U + PV

Two Heat Capacities from q

CalorimetryReaction carried out under constant volume.Use a bomb calorimeter.Usually study combustion.Bomb Calorimetry (Constant Volume Calorimetry)

Constant Pressure (Solution) CalorimetryAtmospheric pressure is constant!

Calorimetry

CalorimetryConstant Pressure Calorimetry

Calorimetry ExamplesIn an experiment similar to the procedure set out for Part (A) of the Calorimetry experiment, 1.500 g of Mg(s) was combined with 125.0 mL of 1.0 M HCl. The initial temperature was 25.0oC and the final temperature was 72.3oC. Calculate: (a) the heat involved in the reaction and (b) the enthalpy of reaction in terms of the number of moles of Mg(s) used. Ans: (a) 25.0 kJ (b) 406 kJ/mol50.0 mL of 1.0 M HCl at 25.0oC were mixed with 50.0 mL of 1.0 M NaOH also at 25.0oC in a styrofoam cup calorimeter. After the mixing process, the thermometer reading was at 31.9oC. Calculate the energy involved in the reaction and the enthalpy per moles of hydrogen ions used. Ans: -2.9 kJ , -58 kJ/mol [heat of neutralization for strong acid/base reactions]CyberChem video

Calorimetry Examples: Hints50.0 mL of 1.0 M HCl at 25.0oC were mixed with 50.0 mL of 1.0 M NaOH also at 25.0oC in a styrofoam cup calorimeter. After the mixing process, the thermometer reading was at 31.9oC. Calculate the energy involved in the reaction and the enthalpy per moles of hydrogen ions used. Ans: -2.9 kJ , -58 kJ/mol [heat of neutralization for strong acid/base reactions]

Comparison of Heat Capacities: CP vs. CV

Example on State and Path Functions20.0 grams of Argon gas was heated from 20.0oC to 80.0oC. Considering Argon behaving as an ideal gas, calculate q, w, U, and H for the following processes:under Constant Volume, andunder Constant Pressure.

Constant Volume Constant Pressure q 374 J 623 J q w 0 -249 J w U 374 J 374 J U H 623 J 623 J H CVm12.5 J mol-1 K-120.8 J mol-1 K-1 CPm

Enthalpy of Phase TransitionsConstant P:s g s g

Enthalpy of Phase Transitions

Chart1

2

2.5

6

9

29

30

Temperature (oC)

Heat Flow at Const. P

Enthalpy of Phase Transitions

Sheet1

Enthalpy

PointTemp (oC)Heat (kJ)

A-252

B02.5

C06

D1009

E10029

F12530

Sheet1

Heat Added (kJ)

Temperature (oC)

Phase Transitions of Water

Sheet2

Temperature (oC)

Heat Flow at Const. P

Enthalpy of Phase Transitions

Sheet3

Relating Urxn and Hrxn

Hesss law: if a reaction is carried out in a number of steps, H for the overall reaction is the sum of H for each individual step.For example:CH4(g) + 2O2(g) CO2(g) + 2H2O(g)H = -802 kJ2H2O(g) 2H2O(l) H = -88 kJCH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = -890 kJ

Hesss Law

Hesss LawGiven:(i)Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)H = -26.7 kJ/mol(ii)CO(g) + O2(g) CO2(g)H = -283.0 kJ/mol

Calculate the heat of reaction for:2Fe(s) + 3/2 O2(g) Fe2O3(s)Ans: -822.3 kJ/mol

If 1 mol of compound is formed from its constituent elements (standard state), then the enthalpy change for the reaction is called the enthalpy of formation, Hof .Standard conditions (standard state): Most stable form of the substance at 1 atm and 25.00 oC (298.15 K). Standard enthalpy, Ho, is the enthalpy measured when everything is in its standard state.Standard enthalpy of formation: 1 mol of compound is formed from substances in their standard states.Standard enthalpy of formation of the most stable form of an element is zero.

Enthalpies of Formation

Enthalpies of Formation: ExampleExample: Write the balanced reaction equation for the standard enthalpy of formation of solid ammonium carbonate.

Enthalpies of Formation

SubstanceDHof (kJ/mol)C(s, graphite)0O(g)247.5O2(g)0N2(g)0

Bucky Ball drawn by HyperChem

Using Enthalpies of Formation to Calculate Enthalpies of ReactionFor a reaction

Calculate/Compare heat of reactions for the combustion of methanol gas and ethanol gas giving carbon dioxide and water.Enthalpies of Formation

Temperature Dependence of CP and Hrxn

Temperature Dependence of CP and Hrxn Example: Find the heat of reaction for the following reaction at 1000. K ( Horxn,1000 K ? ) NaCl(s) Na(g) + Cl2(g)[ Given: Horxn,298K = 519.23 kJ ]

SpeciesCP / J K-1 mol-1Na(g)20.80NaCl(s)49.70Cl2(g)31.70 + 10.14x10-3 T 2.72x10-7 T2

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