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Work, Heat and Internal Energy: The First Law
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System – the specific part of the universe of interest to us Surroundings – the part of the universe not contained in the system
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3 types of Systems open system – exchanges mass and energy closed system – exchanges energy but no mass isolated system – no exchange of either mass or energy
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Open system Closed System corkinsulation Isolated System
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State of a system the system is in a definite state when each of its properties has a definite value. Change in state initial state final state Path initial and final states intermediate states
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Process reversible or irreversible transformation Cyclic transformation begins and ends at the same state variables.
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Isothermal dT = 0 Isochoric dV = 0 Isobaric dP = 0
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Work (w) any quantity that flows across the system’s boundary and is completely convertible into the lifting of a mass in the surroundings. How much work was done? Unit of work = J = 1 kg m/s 2
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A single-stage expansion process State 1 State 2 Piston (T, P 1, V 1 ) mass (m) Piston (T, P 2, V 2 ) mass (m) Direction of piston h2h2 h1h1
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The work done in the surroundings w surr = P ext V The work done by the system w sys = - w surr = - P ext V For an infinitesimal volume change dw sys = - P ext dV
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If the system is in equilibrium F sys = -F ext P = P ext For a simple system d w rev = - P dV
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Ideal gas as the working fluid.
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For an isothermal process (ideal gas as working fluid)
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dw irr = -P ext dV for a constant external pressure
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Heat - the quantity that flows across the boundary of the system during a change in state due to temperature difference between system and surroundings HOT to COLD (never the other way around)!!!
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Measured by determining the temperature change of some known object C - the heat capacity of the system.
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Integrate the infinitesimal heat flow
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Exothermic - system to surroundings Endothermic – surroundings to system surroundings system heat
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Heat flows during phase changes - latent heats Latent heat of vapourisation Latent heat of fusion
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Subject our system to a cyclic transformation
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The following would be true for an exact differential
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The infinitesimal change in the internal energy For a general process
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In general, we write U as a function of T and V
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Examine the first partial derivative
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Define the constant volume heat capacity, C V
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For a system undergoing an isochoric temperature change For a macroscopic system
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Examine the second partial derivative
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A T 1, V m,1, P 1 B Stirrer Valve Thermal insulation FF OO C O C O 50 40 30 20 10 0 20 30 40 50 120 100 80 0 20 40 60 40
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The partial derivative is known as the Joule coefficient, J.
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The change in the internal energy under isothermal conditions is related to the Joule Coefficient
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For an adiabatic process, q = 0!! The first law becomes
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For an ideal gas undergoing a reversible, adiabatic process
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Defining the enthalpy of the system Re-examine the piston with the weight on top Piston (T, P, V) mass (m)
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The first law n Integrating
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Define the enthalpy of the system, H
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In general, we write H as a function of T and P
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Examine the first partial derivative
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Define the constant pressure heat capacity, C P
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For a system undergoing an isobaric temperature change For a macroscopic system
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For an ideal gas In general
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Examine the second partial derivative
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Porous Plug Thermal insulation T 1, P 1, V m,1 T 2, P 2, V m,2 FF OO C O C O 50 40 30 20 10 0 20 30 40 50 120 100 80 0 20 40 60 40 FF OO C O C O 50 40 30 20 10 0 20 30 40 50 120 100 80 0 20 40 60 40
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The partial derivative is known as the Joule-Thomson coefficient, JT.
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The change in the enthalpy under constant pressure conditions is related to the Joule-Thomson Coefficient
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The shorthand form for a chemical reaction J = chemical formula for substance J J = stoichiometric coefficient for J
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The enthalpy change for a chemical reaction H m [J] = molar enthalpies of substance J n J = number of moles of J in the reaction
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Reaction beginning and ending with equilibrium or metastable states Note – Initial and final states have the same temperature and pressure!
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We note that 1 mole of a reaction occurs if
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A reaction that begins and ends with all substances in their standard states The degree sign, either or P = 1.00 bar [aqueous species] = 1.00 mol/ kg T = temperature of interest (in data tables - 25 C or 298 K).
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We note that for 1 mole of a reaction under standard conditions
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A "chemical thermodynamic reference point." For CO and CO 2 C (s) + O 2 (g) CO 2 (g) C (s) + ½ O 2 (g) CO (g)
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The formation reaction 1 mole of a compound constituent elements stable state of aggregation at that temperature. Formation of 1.00 mole of Na 2 SO 3 (s) 2 Na(s) + S(s) + 3/2 O 2 (g) Na 2 SO 3 (s) ‘Formation enthalpy of Na 2 SO 3 (s)’, f H°[Na 2 SO 3 (s)]
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f H° is a measurable quantity! Compare CO (g) with CO 2 (g) C (s) + 1/2 O 2 (g) CO (g) f H° [CO(g)] = -110.5 kJ/mole C (s) + O 2 (g) CO 2 (g) f H° [CO 2 (g)] = - 393.5 kJ/mole
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Formation enthalpies - thermodynamic reference point! H o m [J] = f H [J] H m [elements] = 0 kJ / mole. Use the tabulated values of the formation enthalpies
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The enthalpy change for a given reaction is calculated from the formation enthalpies as Notes Reverse a reaction Multiply a reaction by an integer
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A calorimeter - device containing water and/or another substance with a known heat capacity Calorimeters – either truly or approximately adiabatic systems
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U = q v.
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H = q p
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The enthalpy and the internal energy both represent quantities of heat. U = q v. H = q p. Relate the two state functions using the following relationship U = H - PV
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Enthalpy of solution Enthalpy of dilution Enthalpy of fusion Enthalpy of vapourisation
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sol H - heat absorbed or released when a quantity of solute is dissolved in fixed amount of solvent sol H = H m (sol’n) – H m (component) H(component) = H m (solid) + H m (solvent) Two definitions Standard Limiting
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For the process, HCl (aq, 6 M) HCl (aq, 1 M). The Enthalpy of dilution of the acid. dil H = H m (sol’n 2) – H m (sol’n,1)
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Differentiate the reaction enthalpy with temperature
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r C p - the heat capacity change for the reaction
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Examine a chemical reaction. C (s) + O 2 (g) CO 2 (g) U = U[CO 2 (g)] – U[C(s)] – U[O 2 (g)] Note - r H = -393.5 kJ/mole
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Use tabulated values of formation enthalpies to obtain r H°. May also estimate reaction enthalpies using an indirect method.
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Hess’s Law – the enthalpy change for a given reaction is the same whether the reaction occurs in a single step or in many steps.
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Examine the following reactions H 2 (g) H (g) + H (g) U° = 433.9 kJ Cl 2 (g) Cl (g) + Cl (g) U° = 239.5 kJ Bond dissociation energies. Enthalpy changes are designated D (H- H) and D (Cl-Cl).
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CO 2 (g) C (g) + 2 O (g) U = 740 kJ H of this reaction D(C=O) What about dissociating methane into C + 4 H’s? CH 4 (g) C(g) + 4 H(g) U° = 1640 kJ 4 C-H bonds in CH 4 D (C-H) 410 kJ/mol
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Note: all chemical reactions involve the breaking and reforming of chemical bonds Bonds break - we add energy. Bonds form - energy is released. r U° D(bonds broken) - D(bonds formed)
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These are close but not quite exact. Why? The bond energies we use are averaged bond energies ! This is a good approximation for reactions involving diatomic species. Can only use the above procedure for GAS PHASE REACTIONS ONLY!!!
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