Thermochemistry Notes

Thermochemistry is the branch of chemistry that deals with the study of heat change in
chemical reactions.

Heat(q) is thermal transfer of energy between two bodies that are at different temperatures.
Heat flows from a hot body to a cold body.

In any discussion of heat flow, it is important to distinguish between system and surrounding.

      System: the part of the universe that is of interest to the experimenter.
      Surrounding: the part of the universe that is outside the system.
      Universe = system + surrounding

Three types of systems:

Open system     : allows the exchange mass and energy.
Closed system   : allows the transfer of energy (heat) but not mass.
Isolated system : does not allow the transfer of either mass or energy.

Direction and Sign of Heat Flow
   The heat (q) flow for the system may be a positive or a negative quantity.

      q is + when heat flows into the system from the surrounding

      q is – when heat flows out of the system into the surrounding

The reaction in a system may be an endothermic process or an exothermic process.

      Endothermic process (q>0): a process in which heat has to be supplied to the system
      from the surrounding.
      Example: melting of ice absorbs heat from the surrounding
                H2O(s) + heat  H2O(l)

      Exothermic process (q<0):        any process that gives off or releases heat from the
      reaction system to the surrounding.
      Example: combustion of methane
                CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) + heat


      Calorimetry: measurement of heat changes in physical and chemical processes.
      Calorimeter: apparatus used to measure heat changes in the laboratory.

Chem 100 Notes
mgg                                                                                        Page 1
Thermochemistry Notes

      Heat Capacity (C)
        o the amount of heat required to raise the temperature of the system by 1oC
        o common unit: J/oC
        o an extensive property

      Specific heat (s)
        o the amount of heat required to raise the temperature of 1 gram of a substance by
        o common unit: J/g∙oC
        o an intensive property and can be used to identify a substance

The relationship between the heat capacity and specific heat of a substance is given by:

where m= mass of the substance in grams

If we know the specific heat and the amount of substance, then sample’s temperature change,
∆t, will tell us the amount of heat absorbed or produced in a certain chemical process. The
equation of calculating the heat change is given by:

Since C = ms then,

  Unit conversions:       1 cal = 4.184 J
                         1 kJ = 1000 J

Sample Problems:

1] How much heat is given off when one mole of liquid water cools from 100.0 oC to 10.0 oC?
The specific heat of liquid H2O is 4.18 J/g∙oC
                         = 10.0 oC – 100.0 oC = -90.0 oC

Chem 100 Notes
mgg                                                                                        Page 2
Thermochemistry Notes

2] The temperature of an 89.1 g piece of a metal rises from 22.0 o5C to 51.1 oC when the metal
absorbs 794 J of energy. What is the specific heat of metal?
Given: m=89.1 g; q=794 J;        ∆t=tfinal-tinitial = 51.1oC -22.5oC = 29.1oC


Coffee-Cup Calorimeter
    often used in laboratory classes to measure “heats of reaction” at constant pressure, q p, in
      aqueous solution
    consists of a polystyrene foam cup partially filled with water
    has a tightly fitting cover through which an accurate thermometer is inserted.
    In this apparatus: system = reactants and products
                           surrounding= calorimeter plus the solution (mostly water)
    Heat evolved by the reaction is absorbed by the calorimeter and the solution

Other Thermochemistry Concepts

           Enthalpy: property of a substance that can be used to calculate the heat absorbed or
           produced in a chemical reaction.

                                                                                      Sign Of ∆H
Endothermic (heat absorbed)            Reactants + heat  products                        +∆H
Exothermic (heat released)             Reactants  products + heat                        - ∆H

           Enthalpy of Reaction (∆H rxn): the heat evolved or absorbed by a reaction carried out
           at constant pressure.

∆H   rxn   = enthalpy of the system before reaction – enthalpy of the system after reaction

∆H   rxn   = H(products) – H(reactants)

Chem 100 Notes
mgg                                                                                           Page 3
Thermochemistry Notes

    If the process is endothermic; ∆H   rxn   is positive

    If the process is exothermic; ∆H    rxn   is negative

      Thermochemical equation : a balanced equation that shows both mass and enthalpy

Examples: CaCO3(s)  CaO(s) + CO2(g)                               ∆H   rxn=   178 kJ
          2 SO2(g) + O2(g)  2 SO3(g)                              ∆H   rxn   = -198 kJ

There are several points to keep in mind when interpreting thermochemical equations.

1] If we reverse a thermochemical equation, the magnitude of ∆H is the same, but its sign
       CaCO3(s)  CaO(s) + CO2(g)                                 ∆H rxn= 178 kJ

But the reverse equation must release the same amount of heat.
       CaO(s) + CO2(g)  CaCO3(s)                                  ∆H   rxn=   -178 kJ

2] If we multiply both sides of a thermochemical equation by a factor n, then ∆H must also
change by the same factor.
       [ CaCO3(s)  CaO(s) + CO2(g) ] x 2                           [ ∆H rxn= 178 kJ ] x 2

       2CaCO3(s)  2CaO(s) + 2CO2(g)                               ∆Hrxn = 356 kJ

3] Always specify the physical states of all reactants and products, because they help determine
the magnitude of the enthalpy change.

Sample Problems:

1] Given the thermochemical equation

                                                                 ∆H rxn = -99.1 kJ/mol
  Calculate the heat evolved when 74.6 g SO2 (MM= 64.07 g/mol) is converted to SO3?

Chem 100 Notes
mgg                                                                                          Page 4
Thermochemistry Notes

2] Calculate the heat evolved when 266 g of white phosphorus (P4) burns in air according to the
                                                                          ∆H   rxn=-3013   kJ


    Standard Enthalpy of Formation is the heat released or absorbed when 1 mole of a
    compound is formed from its elements in their standard states at a pressure of 1 atm.

The standard enthalpy of formation of elements in their standard states is 0 kJ.

What is a standard state?
  For a compound
  1) In gaseous state
               Standard state is when the pressure of the gas is 1 atm
  2) In pure liquid or solid
               Standard state is when the solid or liquid is pure

    For an element
       The standard state is the form in which the element exists at 1 atm and 25oC

      Standard Enthalpy of Reaction is the enthalpy of a reaction carried out at 1 atm.
      Consider the hypothetical reaction

Where m and n denote the stoichiometric coefficients for the reactants and products and the
Σ (sigma) means “the sum of”.

To calculate for the     , we must know the      values of the compounds that take part in
the reaction. The values may be determined by applying the direct or the indirect method.

Chem 100 Notes
mgg                                                                                             Page 5
Thermochemistry Notes

      DIRECT METHOD: for compounds that can be readily synthesized from their elements.


1] Calculate the        for C4H10 from the following reaction

      2C4H10(g) + 13O2 (g)  8CO2(g) + 10H2O (g)



2] Benzene (C6H6) burns in air to produce carbon dioxide and liquid water according to the

                   2C6H6(l) + 15O2(g)  12CO2(g) + 6H2O (l)

 Calculate the heat released (in kJ) per gram of the compound reacted with oxygen.    The
     (C6H6,l) = 49.04 kJ /mol.


Chem 100 Notes
mgg                                                                                 Page 6
Thermochemistry Notes

        INDIRECT METHOD: for compounds that cannot be readily synthesized from their
        elements;  can be determined based on Hess’s law of heat summation.

Hess’s Law: The overall ∆H for a reaction equals the sum of all ∆H’s for the individual steps

Using   Hess’s Law:
   1.   Work backward from the final equation
   2.   Reverse reactions as needed, being sure to reverse also ∆H.
   3.   Remember that identical substances found on both sides of the equation cancel each


1] Gasoline, which contains octane as one component, may burn to carbon monoxide if the air
supplied is restricted. Derive the standard reaction enthalpy for the incomplete combustion of
                      2 C8H18(l) + 17O2(g)  16CO(g) + 18H2O(l)

from the standard reaction enthalpies for the combustion of octane and carbon monoxide:

(1) 2 C8H18(l) + 25 O2(g)  16 CO2 (g) + 18 H2O(l)                             ∆Ho = -10 942 kJ
(2) 2 CO(g) + O2(g)  2 CO2(g)                                                 ∆Ho= -566.0 kJ

In the first reaction, the reactant is already on the left side of the equation so the reaction is left

      2 C8H18(l) + 25 O2(g)  16 CO2 (g) + 18 H2O(l)                          ∆Ho = -10 942 kJ

In the second reaction, CO must be on the product side so we need to reverse the equation and
change the sign of its enthalpy and multiply it by 8 to give the coefficients required in the final

      [2 CO2(g)  2 CO(g) + O2(g)] x 8                                        [∆Ho= 566.0 kJ] x 8

      16 CO2(g)  16 CO(g) + 8 O2(g)                                          ∆Ho= 4528 kJ

Add the two chemical equations and their corresponding ∆H.

     2 C8H18(l) + 25 O2(g)  16 CO2 (g) + 18 H2O(l)                     ∆Ho = -10 942 kJ
     16 CO2(g)  16 CO(g) + 8 O2(g)                                           ∆Ho= 4528 kJ

Chem 100 Notes
mgg                                                                                              Page 7
Thermochemistry Notes

    2 C8H18(l) + 17 O2(g)  16 CO(g) + 18 H2O(l)                          ∆Ho= - 6414 kJ

2] Given the following reactions:

S(s) + O2  SO2(g)
H2(g) + ½ O2(g)  H2O(l)
H2S(g) +     O2 (g)  SO2(g) + H2O (l)

Calculate the     of H2S
                          S(s) + H2(g)  H2S(g)
These equations can be arranged so that their sum is the formation reaction of hydrogen sulfide.

The first one is unchanged:  S(s) + O2  SO2(g)
The second one is unchanged: H2(g) + ½ O2(g)  H2O(l)
The third one is reversed:   SO2(g) + H2O (l) H2S(g) +          O2 (g)

The sum is the overall equation: S(s) + H2(g)  H2S(g)                          = -20.2 kJ

Chem 100 Notes
mgg                                                                                          Page 8
Thermochemistry Notes

Chem 100 Notes
mgg                     Page 9

To top