Thermodynamics Mr. Leavings

Objectives Use the laws of thermodynamics to solve problems, identify energy flow within a system, determine the classification of various reactions

Thermodynamic Laws 0-(zeroeth law): temperature makes sense! 1-(1 st Law): The energy in the universe is constant (cant be created or destroyed!) 2-(2 nd Law): The entropy (S) of the universe is increasing 3-(3 rd Law): Entropy (S) of a perfect crystal at 0K is Zero

Entropy (S) Molar entropy is the amount of randomness or disorder in 1 mol of a substance. Units – J/Kmol Entropy always increases with temperature Entropy of a gas > liquid > solid Atoms in a gas have the most freedom of movement

Enthalpy (H) Enthalpy is the total energy of a system, like heat or thermal energy Molar enthalpy depends on temperature Quantities needed –Molar Heat Capacity (C) –Change in Temperature (  T) ΔQ = mC(ΔT)

Heat exchange accompanies chemical reactions. Exothermic: Heat flows out of the system (to the surroundings). Endothermic: Heat flows into the system (from the surroundings). Exo and Endothermic

Exothermic

Endothermic

Calorimeters

Bond Energy Bond energy values can be used to calculate approximate energies for reactions. ΔH = ∑(Bonds Broken) - ∑(Bonds Created)

C + O 2  CO 2 Energy ReactantsProducts  C + O 2 C O kJ kJ/mol

12 CaCO 3  CaO + CO 2 Energy ReactantsProducts  CaCO 3 CaO + CO kJ CaCO kJ  CaO + CO 2

13 In terms of bonds C O O C O O Breaking this bond will require energy. C O O O O C Making these bonds gives you energy. In this case making the bonds gives you more energy than breaking them.

Bond Energy

Gibb’s Free Energy (G)  G =  H – T  S Describes spontaneity Negative  G  Reaction will occur on its own Positive  G  Reaction will not occur on its own

Gibbs Free Energy There are three possible conditions: If  G < 0 then the forward reaction is spontaneous. If  G > 0 then the forward reaction is not spontaneous. (However, the reverse reaction is spontaneous.) If  G > 0, work must be supplied from the surroundings to drive the reaction. If  G = 0 then reaction is at equilibrium and no net reaction will occur.

Temperature and sign of  G - + -Spontaneous at all temperatures + - +Nonspontaneous at any temperature Spontaneous at low temp. but will + be nonspontaneous at high temp Nonspontaneous at low temp. but - will be spontaneous at high temp. Sign of  H  S  GTemperature effect  H - T  S =  G

Calculation of  G o We can calculate  G o values from  H o and S o values at a constant temperature of 25ºC and pressure of 1 atm.Example. Determine  G o for the following reaction at 25 o C Equation N 2 (g) + 3H 2 (g) 2NH 3 (g)  H f o, kJ/mol S o, J/K. mol

Calculation of  G o  H o =  n p  H f o products -  n r  H f o reactants = 2 mol NH 3 ( kJ / mol NH 3 ) = kJ  S o =  n p S o products -  n r S o reactants = 2 mol NH 3 ( J/K. mol NH 3 ) - [1 mol N 2 (191.5 J/K. mol N 2 ) + 3 mol H 2 ( J/K. mol H 2 ) ] = J/K

Calculation of  G o  G =  H - T  S kJ - (298.2K) ( kJ/K) = kJ Note. It was necessary to convert  S from units of J/K to kJ/K.  G f o, kJ/mol, = kJ / 2 mol NH 3 = kJ/mol NH 3 This reaction would occur spontaneously under standard conditions at 25 o C.