Chapter 19: Thermodynamics and Equilibrium Chemistry 1062: Principles of Chemistry II Andy Aspaas, Instructor

First Law of Thermodynamics First Law: conservation of energy applied to thermodynamic systems Internal Energy (U): sum of kinetic and potential energies in a system (energy of motion, and energy contained in chemical bonds and intermolecular forces) U is a state function –Changing from one state to another will give you the change in internal energy, and this is independent of path (ΔU = U f – U i is independent of path) Changes of U are caused by exchanges of energy between system and surroundings –As either heat (due to temp change) or work (due to a force moving an object a certain distance)

Enthalpy Change in enthalpy (ΔH) = heat of reaction at constant pressure (q p ) More specifically, H = U + PV ΔH for a reaction can be calculated by summing standard ΔH f values for products and subtracting summed ΔH f values for reactants

Entropy Entropy (S): measure of how dispersed the energy in a system is among all possible ways a system can contain energy –As a coffee cup cools, it heats the surroundings and disperses its energy. S increases –If a gas is allowed to enter an empty chamber, the kinetic energy of the gas molecules distributes through the whole volume: S increases.

Entropy changes ΔS = S f – S i For H 2 O(s)  H 2 O(l), ΔS = 22 J/K Second Law of Thermodynamics –Total entropy for a system and surroundings increases in a spontaneous process Entropy associated with heat flow = q / T Total entropy change for the system = entropy created plus entropy associated with heat flow (ΔS = S created + q / T) Since entropy must be created in a spontaneous process, if ΔS > q / T the process must be spontaneous At equilibrium, no new entropy is being created, so any entropy change is just due to heat flow ΔS = q / T in a system at equilibrium

Enthalpy, entropy, and spontaneity Since ΔH = q p, ΔS > ΔH / T for any spontaneous process (at constant temp and pressure) ΔH – TΔS < 0 for a spontaneous process > 0 for a nonspontaneous process = 0 for a system at equilibrium

3rd Law of Thermodynamics 3rd Law: a perfectly crystalline solid at 0 K has entropy of zero When heat is transferred, ΔS = q / T S o = Standard entropy –Calculated by changing the temperature slightly and measuring the heat absorbed

Predicting changes in entropy Three occurrences will generally increase entropy –A single molecule is broken into 2 or more molecules –The number of moles of gas increases –Phase changes: (s)  (l) or (g); or (l)  (g) An increase in entropy means the sign of ΔS is (+) The opposite of any of the above corresponds with a decrease in entropy, ΔS = (-)

Predicting the sign of ΔS CS 2 (l)  CS 2 (g) 2Hg (l) + O 2 (g)  2HgO (s) CaCO 3 (s)  CaO (s) + CO 2 (g)

Calculating ΔS o ΔS o =  nS o (products) −  mS o (reactants) S o CS 2 (l) = J/K S o CS 2 (g) = J/K

Free energy Gibbs Free Energy, G = H − TS Free energy change, ΔG = ΔH − TΔS ΔG = (−) for a spontaneous process ΔG = (+) for a nonspontaneous process ΔG = 0 for a process at equilibrium

Standard free energy change ΔG o = ΔH o − TΔS o Calculate ΔG o for the following reaction, and predict its spontaneity CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O (g) ΔH f o (kJ) S o (J/K)

Standard free energies of formation ΔG f o = ΔG for production of 1 mol of a compound under standard conditions from elements in their standard forms ΔG o =  n ΔG f o (products) −  m ΔG f o (reactants)

More precise predictions of spontaneity If ΔG o < -10 kJ, the reaction is spontaneous as written, and reactants transform nearly entirely to products when equilibrium is reached If ΔG o > +10 kJ, the reaction is nonspontaneous as written, and the reactants will not give a significant amount of products at equilibrium If ΔG o is between -10 kJ and +10 kJ, an equilibrium mixture of reactants and products will be obtained, with significant amounts of each

Relating ΔG o to the equilibrium constant ΔG o can be converted to ΔG (for nonstandard temperatures) ΔG = ΔG o + RT ln Q (where Q is the reaction quotient) At equilibrium, ΔG = 0, and Q = K ΔG o = -RT ln K –(this equation relates standard free energy change for a reaction with that reaction’s equilibrium constant)

Calculating K from ΔG o 2NH 3 (g) + CO 2 (g) = NH 2 CONH 2 (aq) + H 2 O(l) ΔG o = kJ R = 8.31 J/(K mol) ΔG o = -RT ln K When K > 1, ΔG o < 0 When K 0

Change of free energy with temperature ΔG T o = ΔH o −TΔS o (assuming ΔH o and ΔS o are constant with respect to temperature) ΔHoΔHo ΔSoΔSo ΔGoΔGo Description -+-Spont. at all T +-+Nonspont at all T --+ or -Spont at low T, nonspont at high T +++ or -Nonspont at low T, spont at high T

Example Ba(OH) 2 · 8H 2 O(s) + 2NH 4 NO 3 (s)  Ba(NO 3 ) 2 (aq) + 2NH 3 (g) + 10H 2 O(l) Predict sign of ΔS ΔH o = ΔS o = ΔG o at room temperature = At what temperature does this reaction switch from being spontaneous to nonspontaneous?