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HESS’ LAW, HEATS OF FORMATION

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Presentation on theme: "HESS’ LAW, HEATS OF FORMATION"— Presentation transcript:

1 HESS’ LAW, HEATS OF FORMATION
THERMOCHEMISTRY HESS’ LAW, HEATS OF FORMATION AND PHASE CHANGES

2 CALCULATING HEATS OF RXNS
There are three ways to calculate the energy of a reaction. DH=mCDT (takes a temperature change, mass, and specific heat constant to calculate a ΔHrxn; uses conservation of energy) Enthalpy of Formation (takes data from a table and uses it to calculate the energy of a reaction) Hess’s Law (allows us to take two or more chem rxns with known ΔHrxn and combine them in such a way to calculate the enthalpy of a target reaction)

3 DHrxn = ∑Hproducts - ∑Hreactants
HEATS OF FORMATION Another method of calculating the enthalpy of a reaction is by using heats of formation. There are tables of DHform that we can gather information from Elements are always 0 DHform is dependent on the number of moles We also need to use the equation presented earlier: DHrxn = ∑Hproducts - ∑Hreactants

4 DHrxn = ∑Hproducts - ∑Hreactants
HEATS OF FORMATION Calculate DH for the following reaction: 8 Al(s) + 3 Fe3O4(s)  4 Al2O3(s) + 9 Fe(s) 8(0) 3( ) 4( ) 9(0) DHrxn = ∑Hproducts - ∑Hreactants DHrxn= {4( )+9(0)} – {8(0)+3( )} DHrxn= (-6,702.8) – ( ) DHrxn= -3,347.6 kJ

5 Classwork CH4 + 2O2  CO2 + 2H2O 2C2H6 + 7O2  4CO2 + 6H2O
Use heats of formations calculations to determine the combustion of which hydro-carbon will produce the most energy per mole… (CH4= kJ/mol; C2H6= kJ/mol; C3H8= ; C4H10= kJ/mol) CH4 + 2O2  CO2 + 2H2O 2C2H6 + 7O2  4CO2 + 6H2O C3H8 + 5O2  3CO2 + 4H2O 2C4H O2  8CO2 + 10H2O

6 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 path gets the desired results and either path requires the same amount of heat energy, this is Hess’s Law.

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

8 The key is to keep our eye on the prize, the goal reaction
During any Hess’s Law calculation, there are two things that we are allowed to do to the given reactions in order to manipulate the. We can reverse the reaction in order to make the products reactants, as long as we change the sign of the enthalpy We can also increase or decrease the amounts of reactants or products by multiplying by a factor, as long as we multiply the enthalpy by the same factor The key is to keep our eye on the prize, the goal reaction

9 For example, use Hess’s Law to calculate the enthalpy of formation for the following reaction equation: 2 N2(g) + 5 O2(g)  2 N2O5(g) DHf = ? Given the following reaction equations: 2 2NO(g) + O2(g) 2NO2(g) DH°rxn= -114kJ/mol 4NO2(g) + O2(g) 2N2O5(g) DH°rxn= -110kJ/mol 2 N2(g) + O2(g) 2NO(g) DH°rxn= +181kJ/mol 2(-114 kJ)+(-110 kJ)+2(181 kJ) = 24 kJ

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

11 2 C2H6  2 C2H4 + 2 H2 274kJ/mol 2H2O2H2+O2 484kJ/mol
Example 2: Rearranging and multiplying: 2 C2H6  2 C2H4 + 2 H kJ/mol 2H2O2H2+O kJ/mol 2C2H4+6O2  4H2O+4CO kJ/mol Find the value of H° for the reaction: 2C2H6 + 7O2  4CO2 + 6H2O

12 2 C2H6  2 C2H4 + 2 H2 274kJ/mol 2H2 + O22H2O - 484kJ/mol
Example 2: Rearranging and multiplying: 2 C2H6  2 C2H4 + 2 H kJ/mol 2H2 + O22H2O kJ/mol 2C2H4+6O2  4H2O+4CO kJ/mol Find the value of H° for the reaction: 2C2H6 + 7O2  4CO2 + 6H2O (274kJ)+(-484kJ)+(-2646kJ) = DHrxn -2856 kJ = DHrxn

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14 Classwork C(s) + H2O(g)  CO(g) + H2(g)
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 = kJ CO(g) + ½O2(g)  CO2(g) H = kJ C(s) + O2(g)  CO2 (g) H = kJ H2(s) + ½O2(g)  H2O(g) H = kJ

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16 PHASE CHANGES & HEAT Energy is required to change the phase of a substance The amount of heat necessary to melt 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 mole of substance Heat of vaporization (Hvap) It takes 40.6 kJ of energy to boil away grams of water.

17 MOLAR HEAT DATA FOR SOME COMMON SUBSTANCES
Hfus Hvap Mercury, Hg 2.29kJ/mol 59.1kJ/mol Ethanol, C2H5OH 5.02kJ/mol 38.6kJ/mol Water, H2O 6.00kJ/mol 40.6kJ/mol Ammonia, NH3 5.65kJ/mol 23.4kJ/mol Helium, He 0.02kJ/mol 0.08kJ/mol Acetone 5.72kJ/mol 29.1kJ/mol Methanol, CH3OH 3.16kJ/mol 35.3kJ/mol

18 boils condenses melts freezes

19 There are 5 distinct sections we can divide the curve into
Ice (solid only) Water & ice (solid & liquid) Water only (liquid only) Water & steam (liquid & gas) Steam only (gas 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

20 If we journey through all of the 5 stages of the heating we have 2 phase changes and 3 increases in temperatures Each stage has its own amnt of energy to absorb or release to make the change necessary The total energy of the entire process can be calculated by combining the energies of each stage

21 mCsolidDT+n(DHfus)+mCliquidDT+DHvap+DHgas
DHtotal = DHsolid +DHmelting +DHliquid +DHvaporizing +DHgas DHtotal = mCsolidDT+n(DHfus)+mCliquidDT+DHvap+DHgas

22 DHtotal = mCsolid DT +n(DHfus) +mCliquidDT DHsolid +DHmelting +DHliquid +n(DHvap) +DHvaporizing +mCgasDT +DHgas

23 SOLID ICE Let’s say we have 180.0g 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 (to the melting point)

24 DHice =(180g)(2.09J/g°C)(0°C-(-10°C))
We can calc the energy absorbed by the ice to this point Use Cice=2.09J/g°C DHice=mCice DT DHice =(180g)(2.09J/g°C)(0°C-(-10°C)) DHice= 3762J

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 KE of the molecules (graph 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 energy required to change from the solid to a liquid is called the heat of fusion & depends on the mols of the substance (DHfus of H2O=6000J/mol or 6kJ/mol) Using the formula: DHmelting = (mol) DHfus 18g H2O 1 mol H2O 1 mol H2O 6000J 180g H2O = 60,000J

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=4.18 J/g°C

28 DHwater=(180g)(4.18J/g°C)(100°C-0°C)
The temp continues to increase until it just reaches the boiling point (for water = 100˚C) again, Hwater=mCwaterT DHwater=(180g)(4.18J/g°C)(100°C-0°C) DHwater= 75,240 J

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=40,600J/mol

30 18g H2O 1 mol H2O 1mol H2O 40,600J 180g H2O = 406,000 J

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=1.84 J/g°C The temp can increase indefinitely, or until the substance decomposes (plasma) We’ll stop at 125°C.

32 DHsteam=(m)(Csteam)(T) DHsteam=(180g)(1.84J/g°C)(125°-100°)
DHsteam= 8280 J

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=(3760 J+60,000J+75,240 J+ 406,000 J+8280 J) DHtotal = 553,280 J Notice, the majority of the energy is needed for the vaporization step. The connections between molecules of H2O must be broken completely to vaporize

34 DHtotal= DHliquid wax + DHsolidification + DHsolid wax
Example: How much energy must be lost for 50.0 g of liquid wax at 85.0˚C to cool to room temperature at 25.0˚C? (Csolid wax= 2.18 J/g˚C, m.p. of wax = 62.0 ˚C, Cliquid wax=2.31 J/g˚C; MM = g/mol, DHfusion=70,500 J/mol) DHtotal= (50g)(2.31J/g˚C)(62˚C-85˚C) + (50g/352.7g/mol)(-70,500J/mol)+ (50g)(2.18J/g˚C)(25˚C-62˚C) DHliquid wax mCliquid waxDT n(DHfusion) DHsolidification DHtotal= ( J) + ( J)+ (-4033 J) DHsolid wax mCsolid waxDT DHtotal= -16,683.8 J DHtotal= DHliquid wax + DHsolidification + DHsolid wax DHtotal= mCliquid waxDT+n(DHfusion)+ mCsolid waxDT

35 Classwork We have a collection of steam at 173°C that occupies a volume of 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?


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