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CALCULATING HEATS OF RXNS  There are three ways to calculate the energy of a reaction. –  H=mC  T (takes a temperature change, mass, and specific.

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Presentation on theme: "CALCULATING HEATS OF RXNS  There are three ways to calculate the energy of a reaction. –  H=mC  T (takes a temperature change, mass, and specific."— Presentation transcript:

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2 CALCULATING HEATS OF RXNS  There are three ways to calculate the energy of a reaction. –  H=mC  T (takes a temperature change, mass, and specific heat constant to calculate a ΔH rxn ; 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 ΔH rxn 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  H form that we can gather information from o Elements are always 0 o  H form is dependent on the number of moles – We also need to use the equation presented earlier:  H rxn = ∑H products - ∑H reactants

4 HEATS OF FORMATION Calculate  H for the following reaction: 8 Al(s) + 3 Fe 3 O 4 (s)  4 Al 2 O 3 (s) + 9 Fe(s) 8(0)3( )4( )9(0)  H rxn = ∑H products - ∑H reactants  H rxn = {4( )+9(0)} – {8(0)+3( )}  H rxn = ( -6,702.8) – ( )  H rxn = -3,347.6 kJ

5 Classwork CH 4 + 2O 2  CO 2 + 2H 2 O 2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 2C 4 H O 2  8CO H 2 O Use heats of formations calculations to determine the combustion of which hydro- carbon will produce the most energy per mole… (CH 4 = kJ/mol; C 2 H 6 = kJ/mol; C 3 H 8 = ; C 4 H 10 = kJ/mol)

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  H sublimation by combining the  H fus with the  H vap 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 2 N 2(g) + 5 O 2(g)  2 N 2 O 5(g)  H  f = ?  For example, use Hess’s Law to calculate the enthalpy of formation for the following reaction equation: 2NO (g) + O 2(g)  2NO 2(g)  H° rxn = -114kJ/mol 4NO 2(g) + O 2(g)  2N 2 O 5(g)  H° rxn = -110kJ/mol N 2(g) + O 2(g)  2NO (g)  H° rxn = +181kJ/mol  Given the following reaction equations: 2 2 2(-114 kJ)+(-110 kJ)+2(181 kJ) = 24 kJ

10 2H 2 O  2H 2 +O 2 484kJ/mol  Example 2: Given the following information: 2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O C 2 H 6  C 2 H 4 + H 2 137kJ/mol 2H 2 O+2CO 2  C 2 H 4 +3O kJ/mol Find the value of  H° for the reaction:

11 2H 2 O  2H 2 +O 2 484kJ/mol  Example 2: Rearranging and multiplying: 2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O 2 C 2 H 6  2 C 2 H H 2 274kJ/mol 2C 2 H 4 +6O 2  4H 2 O+4CO kJ/mol Find the value of  H° for the reaction:

12 2H 2 + O 2  2H 2 O - 484kJ/mol  Example 2: Rearranging and multiplying: 2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O 2 C 2 H 6  2 C 2 H H 2 274kJ/mol 2C 2 H 4 +6O 2  4H 2 O+4CO kJ/mol Find the value of  H° for the reaction: (274kJ)+(-484kJ)+(-2646kJ) =  H rxn kJ =  H rxn

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14 Classwork C (s) + ½O 2(g)  CO (g)  H  = kJ C(s) + H 2 O(g)  CO(g) + H 2 (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. CO (g) + ½O 2(g)  CO 2(g)  H  = kJ C (s) + O 2(g)  CO 2 (g)  H  = kJ H 2(s) + ½O 2(g)  H 2 O (g)  H  = kJ Calculate  H  for this reaction from the following information..

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16 PHASE CHANGES & HEAT PHASE CHANGES & HEAT  Energy is required to change the phase of a substance –The amount of heat necessary to melt 1 mole of substance oHeat of fusion (  H fus ) oIt 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 oHeat of vaporization (  H vap ) oIt takes 40.6 kJ of energy to boil away 18 grams of water.

17 MOLAR HEAT DATA FOR SOME COMMON SUBSTANCES SUBSTANCE  H fus  H vap Mercury, Hg 2.29kJ/mol59.1kJ/mol Ethanol, C 2 H 5 OH 5.02kJ/mol38.6kJ/mol Water, H 2 O 6.00kJ/mol40.6kJ/mol Ammonia, NH kJ/mol23.4kJ/mol Helium, He 0.02kJ/mol0.08kJ/mol Acetone5.72kJ/mol29.1kJ/mol Methanol, CH 3 OH 3.16kJ/mol35.3kJ/mol

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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)  H fus or (#mols)  H vap

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  H total = mC solid  T+n(  H fus )+mC liquid  T+  H vap +  H gas  H total =  H solid  H melting  H liquid  H vaporizing  H gas

22  H total =  H solid  H melting  H liquid  H vaporizing  H gas mC solid  T +n(  H fus ) +mC liquid  T +n(  H vap ) +mC gas  T

23  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) SOLID ICE SOLID ICE

24  H ice =(180g)(2.09J/g°C)(0°C-(-10°C))  H ice = 3762J  H ice =mC ice  T  We can calc the energy absorbed by the ice to this point – Use C ice =2.09J/g°C

25  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 WATER & ICE (MELTING) WATER & ICE (MELTING)  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 (  H fus of H 2 O=6000J/mol or 6kJ/mol)  Using the formula:  H melting = (mol)  H fus 180g H 2 O 1 mol H 2 O 6000J 18g H 2 O 1 mol H 2 O = 60,000J

27  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 – C water =4.18 J/g°C ALL WATER ALL WATER

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

29  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 STEAM & WATER (VAPORIZING)  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  H boiling =(mol)  H vap – H vap of H 2 O=40,600J/mol

30 180g H 2 O 18g H 2 O 1 mol H 2 O 40,600J = 406,000 J

31 STEAM ONLY (VAPOR PHASE) STEAM ONLY (VAPOR PHASE)  Again the heat energy goes into the vibrational energy of the molecule. – Rate of temp increase depends on C H 2 O vapor =1.84 J/g°C  The temp can increase indefinitely, or until the substance decomposes (plasma) – We’ll stop at 125°C.

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

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.  Notice, the majority of the energy is needed for the vaporization step. – The connections between molecules of H 2 O must be broken completely to vaporize  H total =(3760 J+60,000J+75,240 J+ 406,000 J+8280 J)  H total = 553,280 J

34 mC liquid wax  T  H total = (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) 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? (C solid wax = 2.18 J/g˚C, m.p. of wax = 62.0 ˚C, C liquid wax =2.31 J/g˚C; MM = g/mol, DH fusion =70,500 J/mol) Example:  H liquid wax  H solidification  H solid wax  H total=  H liquid wax +  H solidification +  H solid wax n(  H fusion ) mC solid wax  T  H total= mC liquid wax  T+n(  H fusion )+ mC solid wax  T  H total = ( J) + ( J)+ (-4033 J)  H total = -16,683.8 J

35 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? Classwork


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