Work, Heat and Internal Energy: The First Law

System – the specific part of the universe of interest to us Surroundings – the part of the universe not contained in the system

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

Open system Closed System corkinsulation Isolated System

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

Process reversible or irreversible transformation Cyclic transformation begins and ends at the same state variables.

Isothermal dT = 0 Isochoric dV = 0 Isobaric dP = 0

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

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

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

If the system is in equilibrium F sys = -F ext P = P ext For a simple system d w rev = - P dV

Ideal gas as the working fluid.

For an isothermal process (ideal gas as working fluid)

dw irr = -P ext dV for a constant external pressure

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)!!!

Measured by determining the temperature change of some known object C - the heat capacity of the system.

Integrate the infinitesimal heat flow

Exothermic - system to surroundings Endothermic – surroundings to system surroundings system heat

Heat flows during phase changes - latent heats Latent heat of vapourisation Latent heat of fusion

Subject our system to a cyclic transformation

The following would be true for an exact differential

The infinitesimal change in the internal energy  For a general process

In general, we write U as a function of T and V

Examine the first partial derivative

Define the constant volume heat capacity, C V

For a system undergoing an isochoric temperature change  For a macroscopic system

Examine the second partial derivative

A T 1, V m,1, P 1 B Stirrer Valve Thermal insulation FF OO C O C O

The partial derivative is known as the Joule coefficient,  J.

The change in the internal energy under isothermal conditions is related to the Joule Coefficient

For an adiabatic process, q = 0!! The first law becomes

For an ideal gas undergoing a reversible, adiabatic process

Defining the enthalpy of the system Re-examine the piston with the weight on top Piston (T, P, V) mass (m)

The first law n Integrating

Define the enthalpy of the system, H

In general, we write H as a function of T and P

Examine the first partial derivative

Define the constant pressure heat capacity, C P

For a system undergoing an isobaric temperature change  For a macroscopic system

For an ideal gas In general

Examine the second partial derivative

Porous Plug Thermal insulation T 1, P 1, V m,1 T 2, P 2, V m,2 FF OO C O C O FF OO C O C O

The partial derivative is known as the Joule-Thomson coefficient,  JT.

The change in the enthalpy under constant pressure conditions is related to the Joule-Thomson Coefficient

The shorthand form for a chemical reaction  J = chemical formula for substance J J = stoichiometric coefficient for J

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

Reaction beginning and ending with equilibrium or metastable states Note – Initial and final states have the same temperature and pressure!

We note that 1 mole of a reaction occurs if

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).

We note that for 1 mole of a reaction under standard conditions

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)

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)]

 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)] = kJ/mole C (s) + O 2 (g)  CO 2 (g)  f H° [CO 2 (g)] = kJ/mole

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

The enthalpy change for a given reaction is calculated from the formation enthalpies as Notes Reverse a reaction Multiply a reaction by an integer

A calorimeter - device containing water and/or another substance with a known heat capacity Calorimeters – either truly or approximately adiabatic systems

 U = q v.

 H = q p

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

Enthalpy of solution Enthalpy of dilution Enthalpy of fusion Enthalpy of vapourisation

 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

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)

Differentiate the reaction enthalpy with temperature

 r C  p - the heat capacity change for the reaction

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  = kJ/mole

Use tabulated values of formation enthalpies to obtain  r H°. May also estimate reaction enthalpies using an indirect method.

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.

Examine the following reactions H 2 (g)  H (g) + H (g)  U° = kJ Cl 2 (g)  Cl (g) + Cl (g)  U° = kJ Bond dissociation energies. Enthalpy changes are designated D (H- H) and D (Cl-Cl).

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

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)

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!!!