Thermochemistry
Advertisement
This presentation is the property of its rightful owner.
1 / 17

Thermochemistry PowerPoint PPT Presentation

Thermochemistry. Calorimetry Example If 10.0 g of Al metal is heated to 75.00 C and transferred to a coffee cup calorimeter containing 95.0 g H 2 O at 25.00 C , what is the specific heat of Al if the contents of the calorimeter end up at 26.11 C?

Download Presentation

Thermochemistry

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


Lecture 11 62800

Thermochemistry

Calorimetry ExampleIf 10.0 g of Al metal is heated to 75.00 C and transferred to a coffee cup calorimeter containing 95.0 g H2O at 25.00 C , what is the specific heat of Al if the contents of the calorimeter end up at 26.11 C?

Calorimetry Example: try this one on your own!

If 25.0 g Fe at 95.00 C were added to a coffee cup calorimeter containing 75.0 g H2O at 28.00 C, what would be the final temperature of the contents of the calorimeter when equilibrium is established?

Lecture 11


Lecture 11 62800

Energy

  • Enthalpy (H) is the heat energy associated with systems at constant pressure

    • Most systems studied by chemists are at constant pressure

    • H= Hfinal-Hinitial =qPthe enthalpy change accompanying a change in a system is the amount of heat gained or lost by a system when the change occurs atconstant pressure

    • Enthalpy is a state function

    • H > 0 when enthalpy (qP) is gained by the system

      • This is an endothermic process

    • H < 0 when enthalpy (qP) is lost by the system

      • This is an exothermic process

  • Relationship between E and H

  • E = H - PV

  • at constant pressure

  • E = H - PV

  • for many processes at constant pressure, Vis small

Lecture 11


Lecture 11 62800

Thermochemistry

  • Enthalpies of Reaction are the changes in enthalpy accompanying a chemical reaction

    • Thermochemical equations include the usual balanced equation with the enthalpy change appended to the right of the products

      • Example: 2H2(g) + O2(g) 2H2O(g) DH = -483 kJ

        • DHrxn = H(products) - H(reactants)

        • if DHrxn < 0, H(products) < H(reactants) and enthalpy is lost from the system to the surroundings

        • DHrxn < 0 means the reaction is exothermic

        • if DHrxn > 0, H(products) > H(reactants) and enthalpy is added to the systemfrom the surroundings

        • DHrxn > 0 means the reaction is endothermic

Lecture 11


Lecture 11 62800

Thermochemistry

  • Thermochemical Equations

    • Enthalpy is an extensive property and it is understood that the equation is written on a mole basis

      • 483 kJ are produced for 2 mol H2(g), 1 mol O2(g), 2 mol H2O(g)

      • 2H2(g) + O2(g) 2H2O(g) DH = -483 kJ

      • 4H2(g) + 2O2(g) 4H2O(g) DH = (-483 kJ) x 2 = -966 kJ

  • If the equation is reversed, the enthalpy change will be the same size but opposite sign

    • 2H2O(g) 2H2(g) + O2(g) DH = 483 kJ

  • The enthalpy change for a reaction depends on the state of the reactants and products and states must be included in the equation

  • 2H2(g) + O2(g) 2H2O(g) DH = -483 kJ

  • 2H2(g) + O2(g) 2H2O(l ) DH = -571 kJ

  • 2H2O(g) 2H2O(l ) DH = - 88 kJ

  • Lecture 11


    Lecture 11 62800

    Thermochemistry

    • Hess’s Law

      • If one considers a reaction taking place in a series of steps, the enthalpy change for the overall reaction will be the sum of the enthalpy changes of the individual steps.

        • This results from the fact that H is a state function.

        • One must consider the physical state of all reactants and products.

      • Example

      • 2H2(g) + O2(g) 2H2O(g) DH = -483 kJ

      • 2H2O(g) 2H2O(l ) DH = - 88 kJ

      • (sum) 2H2(g) + O2(g) 2H2O(l ) DH = -571 kJ

      • Example: From the following thermochemical equations, find the enthalpy change for NO(g) + O2 (g)NO2 (g).

        • N2 (g) + O2 (g) NO (g) DH = 90.37 kJ

        • N2 (g) + O2 (g) NO2 (g) DH = 33.8 kJ

    Lecture 11


    Lecture 11 62800

    Thermochemistry

    • Step 1: write the equation for which the DH is sought

      • NO(g) + O2 (g)NO2 (g) DH = ?

  • Step 2: rearrange the equations for the steps so the correct reactants are on the left and the correct products are on the right.

    • Some equations may be reversed: this changes the sign on DH

    • Some equations may be multiplied by an integer or fraction

    • N2 (g) + O2 (g) NO2 (g) DH = 33.8 kJ

    • NO (g)N2 (g) + O2 (g) DH = -90.37 kJ

    • NO(g) + O2 (g)NO2 (g) DH = ?

  • Step 3: Cancel substances in the equations above the line.

    • Substances not in the final equation should all cancel

    • The coefficients for some substances should cancel to give the coefficient for those substances in the final equation.

      • O2(g) - O2 (g) = O2 (g) on the left above

      • N2 (g) - N2 (g) =0N2 (g) above: N2(g) is not in the final equation

  • Lecture 11


    Lecture 11 62800

    Thermochemistry

    • Step 4: add the DH’s to get the DHrxn for the reaction.

      • DHrxn=33.8 kJ - 90.7 kJ = -56.6 kJ

  • Another Example: From the following equations, calculate the DH for

  • S(s) + O2(g) SO2(g)

  • 2SO2 (g) + O2 (g)2SO3 (g)DH= -196 kJ

  • 2S(s) + 3O2 (g) 2SO3 (g)DH = -790 kJ

  • (2SO3 (g) 2SO2 (g) + O2 (g) DH= +196 kJ) x

  • (2S + 3O2 (g) 2SO3 (g)DH = -790 kJ) x

  • S(s) + O2(g) SO2(g) DH = ?

  • 1SO3 (g) 1SO2 (g) + O2 (g) DH= (+196 kJ) x

    1S + O2 (g) 1SO3 (g)DH = (-790 kJ) x

    S(s) + O2(g) SO2(g) DH = ?

    DHrxn= (+196 kJ) + (-790 kJ) = -297 kJ

    Lecture 11


    Lecture 11 62800

    Thermochemistry

    • Standard Enthalpy of Formation, Hf°, is the enthalpy change associated with the formation of one mole of a compound from its elements in their standard states.

      • The standard state is the most stable physical state the compound or element has at 298.15 K (25 °C) and 1 atmosphere pressure

        • Oxygen exists as O2 gas at 25 °C

        • Carbon exists as solid graphite at 25 °C.

        • Water is H2O(l ) in its standard state.

      • Hf°for elements is defined as zero when the elements are in their standard states.

      • Examples:

        • H2(g) + O2(g) H2O(l ) Hf° = -285.8 kJ/mol

        • 3C(s) + 4H2(g) C3H8(g) Hf° = -103.85 kJ/mol

        • Table 6.2, page 270, as does Appendix L in Kotz & Treichel gives many Hf°’s without balanced chemical equations.

    Lecture 11


    Lecture 11 62800

    Thermochemistry

    • Other kinds of enthalpies

      • Enthalpy of vaporization: the enthalpy change accompanying the conversion of one mol of a substance from liquid to gas

        • H2O(l ) H2O(g) Hvap° = + 44 kJ/mol

      • Enthalpy of fusion: the enthalpy change accompanying the conversion of one mol of substance from solid to liquid.

        • H2O(s) H2O(l ) Hfusion° = 6.008 kJ/mol

      • Enthalpy of combustion: the enthalpy change accompanying the complete combustion of one mole of a substance.

        • C7H8(l ) + 9O2(g) 7CO2(g) + 4H2O(l ) Hcomb° = -3,910 kJ

    Lecture 11


    Lecture 11 62800

    Thermochemistry

    • Applications of enthalpies of formation

      • Calculate enthalpy changes for other reactions

        • Example: calculate the Hrxn° for the combustion of benzene, C6H6(l )

        • 6C(s) + 6O2(g) 6CO2(g) Hf° = 6(-393.5 kJ)

        • 3H2(g) + O2(g) 3H2O(l ) Hf° = 3(-285.8 kJ)

        • C6H6(l ) 6C(s) + 3H2(g) Hf° = (-49.04 kJ)

        • C6H6(l ) + O2(g) 6CO2(g) + 3H2O(l ) Hcomb° = ?

        • Hcomb° = 6(-393.5 kJ) + 3(-285.8 kJ) + (-49.04 kJ) = -3,267.4 kJ

        • Notice: we have added the heats of formation of CO2(g) and H2O(l ), weighted for their number of moles, and subtracted the heat of formation of C6H6(l ).

        • This amounts to adding the enthalpies of formation of the products and subtracting the sum of the enthalpies of formation of reactants to get the heat of reaction. These enthalpies are weighted for their respective numbers of moles.

    Lecture 11


    Lecture 11 62800

    Thermochemistry

    • Calculate enthalpy changes for other reactions (cont.)

      • This can be summarized:

      • Hrxn° = n Hf°(products) -m Hf°(reactants)

      • where  means “the sum of”

      • n is the respective molar coefficient for each product

      • m is the respective molar coefficient for each reactant

      • Example: find Hrxn° for NH4NO3(s) N2(g) + 2H2O(g) + O2(g)

      • Hrxn° = Hf°(N2(g) ) + 2Hfo(H2O(g)) + Hf°(O2(g)) -[Hf°(NH4NO3(s))]

      • Hrxn° = 0 +2 x (-241.8 kJ) + 0 - [-365.6]

      • Hrxn° = -118 kJ

    Lecture 11


    Lecture 11 62800

    Thermochemistry

    • Constant-volume Calorimetry - the use of bombs!

      • A bomb in this case is a sealed container, usually made from steel, in which a reaction can be carried out.

      • The major kinds of reactions carried out in bombs are combustion reactions.

      • It turns out that qV = DE

        • DH can be calculated from DH = DE + PDV, where DV is the change in volume that would occur if the process were carried out at constant pressure, P.

      • Normally, the heat capacity, Ccalorimeter=Cwater+Cbomb, of the calorimeter must be known.

        • Combust a quantity of a substance having a known qV.

        • From Ccalorimeter and DT for a mass of another substance, qV = DE can be calculated.

    Lecture 11


    Lecture 11 62800

    Thermochemistry

    • Constant-volume Calorimetry - the use of bombs!

      • Example: When benzoic acid, C7H6O2(s) is combusted in a bomb calorimeter, it has qv=-26.38 kJ/g. The heat capacity of a bomb is determined by sealing 0.3684 g of benzoic acid in a bomb with excess O2 , immersing the bomb in 855.0 g H2O. The temperature is allowed to come to equilibrium at 22.350 oC and the acid is combusted. After thermal equilibrium, the temperature is measured to be 24.500 oC.

    Lecture 11


    Lecture 11 62800

    Thermochemistry

    • Constant-volume Calorimetry - the use of bombs!

      • Example: The bomb from the previous example is used to determine the heat of combustion of naphthalene, C10H8(s). 0.1447 g of naphthalene is sealed in the bomb with excess O2 and the bomb is immersed in 825.3 g H2O. Combustion causes the temperature of the bomb and water to increase by 1.324 oC. What is the molar heat of combustion of naphthalene?

    Lecture 11


    Lecture 11 62800

    Thermochemistry

    • Constant pressure calorimetry

      • Appropriate for measuring heat changes of reactions in solution

      • Heat measurements give DH = qP

      • Heat gained by the solution = (specific heat of sol’n) x (sol’n mass) x DT

        • When a reaction occurs: qrxn = -qsol’n

        • If solution is a dilute aqueous solution, specific heat = 4.18 J/g

        • Use an adiabatic calorimeter: all heat is contained within the calorimeter

    Lecture 11


    Lecture 11 62800

    Thermochemistry

    • Calorimetry Example

      • A student mixes 50.0 mL of 1.00 M HCl and 50.0 mL of NaOH in a coffee cup calorimeter and finds the mixture goes from 21.00 C to 27.50 C. Calculate the change in enthalpy for the reaction, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100.0 mL, that its density is 1.000g/mL and its specific heat is 4.18 J/g K.

      • HCl(aq) + NaOH(aq) H2O(l ) + NaCl(aq) DH = -5.4 kJ

    Lecture 11


  • Login