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THERMOCHEMISTRY

THERMOCHEMISTRY. HESS’ LAW, HEATS OF FORMATION AND PHASE CHANGES. CALCULATING HEATS OF RXNS. There are three ways to calculate the energy of a reaction. D H= mC D T (takes a temperature change, mass, and specific heat constant to calculate a Δ H rxn ; uses conservation of energy)

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THERMOCHEMISTRY

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  1. 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 chemrxns with known ΔHrxn and combine them in such a way to calculate the enthalpy of a target reaction)

  3. 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. HEATS OF FORMATION Calculate DH for the following reaction: 8 Al(s) + 3 Fe3O4(s)  4 Al2O3(s) + 9 Fe(s) 8(0) 3(-1118.4) 4(-1675.7) 9(0) DHrxn = ∑Hproducts - ∑Hreactants DHrxn= {4(-1675.7)+9(0)} – {8(0)+3(-1118.4)} DHrxn= (-6,702.8) – (-3355.2) DHrxn= -3,347.6 kJ

  5. Classwork Use heats of formations calculations to determine the combustion of which hydro-carbon will produce the most energy per mole… (CH4= -74.81 kJ/mol; C2H6= -84.68 kJ/mol; C3H8= -104.5; C4H10= -126.5 kJ/mol) CH4 + 2O2 CO2 + 2H2O 2C2H6 + 7O2 4CO2 + 6H2O C3H8 + 5O2 3CO2 + 4H2O 2C4H10 + 13O2 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. 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. 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

  11. 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

  12. 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

  13. 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

  14. PHASE CHANGES & HEAT • Energy is required to change the phaseof a substance • The amount of heat necessary to melt 1 mole of substance • Heat of fusion (Hfus) • It takes 6.00 kJof 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 kJof energy to boil away 18 grams of water.

  15. boils condenses melts freezes

  16. 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

  17. 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

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

  19. +n(DHfus) DHtotal = mCsolidDT +mCliquidDT +DHmelting +DHliquid +n(DHvap) +DHvaporizing +DHgas DHsolid +mCgasDT

  20. 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 thevibrational 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)

  21. We can calc the energy absorbed by theiceto this point • Use Cice=2.09J/g°C DHice=mCiceDT DHice=(180g)(2.09J/g°C)(0°C-(-10°C)) DHice= 3762J

  22. WATER & ICE (MELTING) • Any additional heat absorbed by the ice goes into partially breaking the connectionsbetween 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

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

  24. 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

  25. 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

  26. 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 thetemp 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 heatof vaporization; using Hboiling=(mol)Hvap • Hvap of H2O=40,600J/mol

  27. 40,600J 1 mol H2O 1mol H2O 18g H2O 180g H2O = 406,000 J

  28. 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.

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

  30. 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

  31. 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 = 352.7 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 mCliquidwaxDT n(DHfusion) DHsolidification DHtotal=(-2656.5 J) + (-9994.3 J)+ (-4033 J) mCsolidwaxDT DHsolid wax DHtotal=-16,683.8 J DHtotal= DHliquid wax + DHsolidification+ DHsolid wax DHtotal= mCliquidwaxDT+n(DHfusion)+ mCsolidwaxDT

  32. 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?

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