Slide1 l.jpg
This presentation is the property of its rightful owner.
Sponsored Links
1 / 35

THERMOCHEMISTRY PowerPoint PPT Presentation

  • Uploaded on
  • Presentation posted in: General

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)

Download Presentation


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Slide1 l.jpg




Slide2 l.jpg


  • 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)

Slide3 l.jpg


  • 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

Slide4 l.jpg


Calculate DH for the following reaction:

8 Al(s) + 3 Fe3O4(s)  4 Al2O3(s) + 9 Fe(s)





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

Slide5 l.jpg


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

Slide6 l.jpg


  • 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.

Slide7 l.jpg

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

Slide8 l.jpg

  • 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

Slide9 l.jpg

  • 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:


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

2(-114 kJ)+(-110 kJ)+2(181 kJ) = 24 kJ

Slide10 l.jpg



  • 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

Slide11 l.jpg

  • Example 2:

    Rearranging and multiplying:

2 C2H6  2 C2H4 + 2 H2 274kJ/mol

2H2O2H2+O2 484kJ/mol


Find the value of H° for the reaction:

2C2H6 + 7O2 4CO2 + 6H2O

Slide12 l.jpg

  • 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

Slide14 l.jpg


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

Slide16 l.jpg


  • 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.

Slide18 l.jpg





Slide19 l.jpg

  • 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

Slide20 l.jpg

  • 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

Slide21 l.jpg

DHtotal =






DHtotal =


Slide22 l.jpg


DHtotal =










Slide23 l.jpg


  • 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)

Slide24 l.jpg

  • We can calc the energy absorbed by theiceto this point

    • Use Cice=2.09J/g°C



DHice= 3762J

Slide25 l.jpg


  • 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

Slide26 l.jpg

  • 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


18g H2O

1 mol H2O

  • Using the formula: DHmelting=(mol) DHfus

180g H2O

= 60,000J

Slide27 l.jpg


  • 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

Slide28 l.jpg

  • The temp continues to increase until it just reaches the boiling point (for water = 100˚C)

    • again, Hwater=mCwaterT


DHwater= 75,240 J

Slide29 l.jpg


  • 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

Slide30 l.jpg


1 mol H2O

1mol H2O

18g H2O

180g H2O

= 406,000 J

Slide31 l.jpg


  • 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.

Slide32 l.jpg



DHsteam= 8280 J

Slide33 l.jpg

  • 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

Slide34 l.jpg


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




DHtotal=(-2656.5 J) + (-9994.3 J)+ (-4033 J)


DHsolid wax

DHtotal=-16,683.8 J

DHtotal= DHliquid wax + DHsolidification+ DHsolid wax

DHtotal= mCliquidwaxDT+n(DHfusion)+


Slide35 l.jpg


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?

  • Login