THERMOCHEMISTRY

1. THERMOCHEMISTRY HESS’ LAW, AND HEAT VS. TEMPERATURE

2. HESS’ LAW • The change in energy of a process or reaction is a state function, meaning that regardless of the path to reach your goal, the energy to get there is constant. • For instance if you want to vaporize a solid, you have two pathways. • You can melt it into a liquid and then vaporize it into a gas. • Or you can sublime the solid directly into a gas. • Either pathway gets the desired results and either pathway requires the same amount of heat energy, this is Hess’s Law.

3. The idea that we can calculate Hsublimation by combining the Hfus with the Hvap is an illustration of Hess’ Law.

4. HESS’ LAW • Hess’ Law is a law of physical chemistry that states the enthalpy change of a reaction is the same regardless of what pathway is taken to achieve products. • Only the start and end states matter to the reaction not the individual steps in between • This allows the H for a rxn to be calculated even when it cannot be measured directly • This process will feel a little like combining half reactions like we did earlier this semester.

5. Only the start and end states matter to the reaction not the individual steps in between • For instance let’s look at this synthesis: • N2 + 2H2O  N2H4 + O2DH = +622.2kJ • 2H2 + O2 2H2ODH= -571.6kJ N2 + 2H2 N2H4 Hrxn: ? • We can calculate Hrxn by using a combination of the following 2 rxns: N2 + 2H2 N2H4 DHrxn = 50.6kJ/mol

6. What if the rxns we are trying to combine don’t combine naturally as presented? • Then we can manipulate the eqnsto make them combine naturally. 2 N2(g) + 5 O2(g) 2 N2O5(g) DHf = ? • We are given a series of related rxns, with enthalpy data, that we will attempt to combine to calculate the Hf above: 2NO(g) + O2(g) 2NO2(g) DH°rxn= -114kJ/mol 4NO2(g) + O2(g) 2N2O5(g) DH°rxn= -110kJ/mol N2(g) + O2(g) 2NO(g) DH°rxn= +181kJ/mol

7. We need to remember that the amount of energy of a reaction is related to both direction (endo/exo or +/-) and moles (coefficients of a rxn equation). • Because of these connections to our enthalpy, any changes that we make to the reaction equation must be reflected in our given H values. • There are two rules to remember during our manipulations: • If we reverse the rxn, the sign of the DH°rxn is reversed. For example, if: If: 2H2(g)+ O2(g) 2H2O (g) DH°rxn= -482kJ/mol Then: 2H2O (g)2H2(g)+ O2(g)DH°rxn= +482kJ/mol

8. If we need to multiply our coefficients in a reaction by a value to increase or decrease the amount of products then our enthalpy value must be multiplied by the same factor. For example if we double the following rxn: If: 2H2(g)+ O2(g) 2H2O (g) DH°rxn= -482 kJ/mol Then: 4H2(g)+2O2(g)4H2O(g) H°rxn= -964 kJ/mol • So let’s use these rules to combine the previous reactions to determine the enthalpy of formation of N2O5: 2 N2(g) + 5 O2(g) 2 N2O5(g) DHrxn = ?

9. 2 N2(g) + 5 O2(g) 2 N2O5(g) DHrxn = ? Goal: • First we want to target reactants or products in our given eqns that already match both in coefficient and in side of eqn with our final reaction eqn. • Since the are identical already, we don’t want to change anything about them. If we were to change something, than they would no longer match. 2NO(g) + O2(g) 2NO2(g) DH°rxn= -114kJ/mol 4NO2(g) + O2(g) 2N2O5(g) DH°rxn= -110kJ/mol N2(g) + O2(g) 2NO(g) DH°rxn= +181kJ/mol

10. 2 N2(g) + 5 O2(g) 2 N2O5(g) DHrxn = ? Goal: • Next we want to target reactants or products in our final rxneqn that are unique to only one of the given rxneqns • We see that in our target eqn we need to end up with 2 mols of N2 gas in the reactants • Our third given eqn has 1 mol of N2 gas in the reactants • If we doubled that rxn it would match… 2NO(g) + O2(g) 2NO2(g) DH°rxn= -114kJ/mol 4NO2(g) + O2(g) 2N2O5(g) DH°rxn= -110kJ/mol N2(g) + O2(g) 2NO(g) DH°rxn= +181kJ/mol 2N2(g) + 2O2(g) 4NO(g) DH°rxn= +362kJ/mol

11. 2 N2(g) + 5 O2(g) 2 N2O5(g) DHrxn = ? Goal: • Now we want to decide how to make use of the first rxn eqn. • If we multiply it by two, we would be able to cancel out the NO2in 1st first eqn and the NO2 in the 2nd eqn. • That manipulation (since we have changed something in each eqn) should set us up to combine the eqns to receive our target eqn. 4NO(g) + 2O2(g) 4NO2(g) DH°rxn= -228kJ/mol 2NO(g) + O2(g) 2NO2(g) DH°rxn= -114kJ/mol 4NO2(g) + O2(g) 2N2O5(g) DH°rxn= -110kJ/mol 2N2(g) + 2O2(g) 4NO(g) DH°rxn= +362kJ/mol

12. 2 N2(g) + 5 O2(g) 2 N2O5(g) DHrxn = ? Goal: • Now we should be able to combine the three into one and end up with the one above 4NO(g) + 2O2(g) 4NO2(g) DH°rxn= -228kJ/mol 4NO2(g) + O2(g) 2N2O5(g) DH°rxn= -110kJ/mol 2N2(g) + 2O2(g) 4NO(g) DH°rxn= +362kJ/mol 2N2(g) + 2O2(g) + O2(g) + 2O2(g)  2N2O5(g) 2N2(g) + 5O2(g)  2N2O5(g) DHrxn1 + DHrxn2 + DHrxn3 = DHrxn (-228 kJ/mol)+(-110 kJ/mol)+(362kJ/mol)= 24kJ

13. 2 2 • Example 2: Given the following information: C2H6C2H4 + H2 137kJ/mol 2H2O2H2+O2 484kJ/mol 2H2O+2CO2C2H4+3O2 1323kJ/mol Find the value of H° for the reaction: 2C2H6 + 7O2 4CO2 + 6H2O

14. Example 2: Rearranging and multiplying: 2 C2H6  2 C2H4 + 2 H2 274kJ/mol 2H2O2H2+O2 484kJ/mol 2C2H4+6O24H2O+4CO2-2646kJ/mol Find the value of H° for the reaction: 2C2H6 + 7O2 4CO2 + 6H2O

15. Example 2: Rearranging and multiplying: 2 C2H6  2 C2H4 + 2 H2 274kJ/mol 2H2 + O22H2O- 484kJ/mol 2C2H4+6O24H2O+4CO2- 2646kJ/mol Find the value of H° for the reaction: 2C2H6 + 7O2 4CO2 + 6H2O (274kJ)+(-484kJ)+(-2646kJ) = DHrxn -2856 kJ = DHrxn

17. Classwork Before pipelines were built to deliver natural gas, individual towns and cities contained plants that produced a fuel known as town gas by passing steam over red-hot charcoal. C(s) + H2O(g)  CO(g) + H2(g) Calculate H for this reaction from the following information.. C(s) + ½O2(g) CO(g)H = -110.53 kJ CO(g) + ½O2(g) CO2(g)H = -282.98 kJ C(s) + O2(g)CO2 (g)H = -393.51 kJ H2(s) + ½O2(g)H2O(g)H = -241.82 kJ

19. PHASE CHANGES & HEAT • Energy is required to change the phase of a substance • The amount of heat necessary to melt 1 mole of substance • Heat of fusion (Hfus) • It takes 6.00 kJ of energy to melt 18 grams of ice into liquid water. • The amount of heat necessary to boil 1 mole of substance • Heat of vaporization (Hvap) • It takes 40.6 kJ of energy to boil away 18 grams of water.

21. boils condenses melts freezes

22. There are 5 distinct sections we can divide the curve into • Ice • Water & ice • Water only • Water & steam • Steam only • We can calculate the amount of energy involved in each stage • There are two types of calculations • Temperature changes use H=mCT • Phase changes use (#mols)Hfus or (#mols)Hvap

23. SOLID ICE • Let’s say we have 18.0g of ice at –10°C, & we begin heating it on a hot plate with sustained continuous heat. • Heat energy absorbs into the ice increas-ing the vibrational or kinetic energy of the ice molecules • The temp will increase & will continue to increase until just before the ice has enough energy to change from solid to liquid • We can calc the energy absorbed by the ice to this point • Use the eqn DH=mCDT (cice=2.09J/g°C)

24. = ice 376.2J H D DHice=mCiceDT DHice =(18g)(2.09J/g°C)(0°C-(-10°C)) DHice=376.2J 0° -10°

25. WATER & ICE (MELTING) • Any additional heat absorbed by the ice goes into partially breaking the connections between the ice molecules. • There is no change in the kinetic energy of the molecules (curve flattens out) • No change in temp • All of the energy goes into breaking the connections • As long there is solid ice present, the temp cannot increase. • The solid & liquid are in equilibrium if they are both present

26. The H required to change from the solid to the liquid phase is called the heat of fusion & depends on the amnt of the substance (DHfus of H2O=6000J/mol or 6kJ/mol) 1 mol H2O 6000J = D H 18g H2O 1 mol H2O = ice 376.2J H melting D 6000J • Using the formula: DHmelting =(mol) DHfus 18g H2O = 6000J 0° -10°

27. ALL WATER • Now all of the particles are free to flow, • The heat energy gained now goes into the vibrational energy of the molecules. • The temp of the water increases • The rate of temp increase now depends on the heat capacity of liquid water • Cwater=4.180J/g°C • The temp continues to increase until it just reaches boiling stage (for water = 100ºC) • again, Hwater=mCwaterT

28. = = D H = ice water 376.2J H melting D H D DHwater=(18g)(4.18J/g°C)(100°C-0°C) DHwater= 7524J 100° 0° 7524J 6000J -10°

29. STEAM & WATER (VAPORIZING) • Any additional heat absorbed by the water goes into completely breaking the connections between the water molecules. • Again the heat does not increase the KE of the molecules so the temp does not change, • the energy is used to vaporize the water • If there are still connections to break or there is liquid present, the temp cannot increase. • The energy required to change from the liquid to the vapor phase is called the heat of vaporization; using Hboiling=(mol)Hvap • Hvap of H2O=40,790J/mol

30. 40,790J 1 mol H2O 1mol H2O 18g H2O 7524J = = D H = ice water 376.2J H melting 40,790J D H D = boiling H D 18g H2O 100° Hboiling=(mol)HVAP 0° 6000J -10°

31. STEAM ONLY (VAPOR PHASE) • Again the heat energy goes into the vibrational energy of the molecule. • Rate of temp increase depends on CH2O vapor=1.84J/g°C • The temp can increase indefinitely, or until the substance decomposes (plasma) • We’ll stop at 125°C.

32. 7524J = = 828J D H = ice water 376.2J H melting 40,790J D H = D steam = H D boiling H D 125° DHSTEAM=(m)(CWATER VAPOR)(T) DH=(18g)(1.84J/g°C)(125°-100°) 100° 0° 6000J -10°

33. To figure out how much energy we need would need all together to heat up the water this much, we just need to add up the energy of each step. DHtotal= (376J + 6000J +7524J + 40,790J + 829J) DHtotal =55,500J • Notice, the majority of the energy is needed for the vaporization step. • The connections between molecules of H2O must be broken completely to vaporize

35. Classwork We have a collection of steam at 173°C that occupies a volume of 30.65 L and a pressure of 2.53 atm. How much energy would it need to lose to end up as a block of ice at 0.00°C?