Module 4

Lesson 3.2  Standard Molar Enthalpies of Formation



Key Concepts


The Table of Standard Molar Enthalpies of Formation is used to:

  • compare stabilities of compounds by predicting standard molar enthalpies of decomposition
  • predict enthalpies of reaction and molar enthalpies of reaction for many chemical reactions
  • predict the standard molar enthalpy of formation for a specified species in a reaction, given a standard enthalpy of reaction


  1. Predicting Standard Molar Enthalpies of Decomposition


    Recall from previous science courses that a simple decomposition reaction is a reaction in which a compound is broken into its constituent elements (Refer to page 58 in the textbook). In other words, simple decomposition reactions are the reverse of formation reactions. For example, the standard molar enthalpy of decomposition of aluminium oxide is represented by the following reaction. Notice that when one mole of aluminium oxide is decomposing into its elements, the sign of the enthalpy is reversed. Although the formation of aluminium oxide releases energy, the decomposition reaction requires energy (that is, it is endothermic).

\( \mathrm { Al_2O_3(s) \rightarrow 2Al(s) + \frac{3}{2}O_2(g) } \) \( \mathrm { \Delta_dH^{\circ} = +1675.7 \frac{kJ}{mol} } \)


This reaction can also be written:


\( \mathrm { Al_2O_3(s) + 1675.7 kJ \rightarrow 2Al(s) + \frac{3}{2}O_2(g) } \)


Thermal stability is the tendency of a compound to resist decomposition when it is heated. Thermal stability allows us to predict how reactive a compound is likely to be. By allowing us to predict quickly the standard molar enthalpies of decomposition, the Table of Standard Molar Enthalpies can be used to compare the thermal stabilities of various compounds. For example, we can use the table to compare the thermal stability of HBr and HCl.


First, use the Table of Standard Molar Enthalpies to find the standard molar enthalpies of formation for the two compounds.

\( \mathrm { \frac{1}{2}H_2(g) + \frac{1}{2}Br_2(l) \rightarrow HBr(g) } \) \( \mathrm { \Delta_fH^{\circ} = -36.3 \frac{kJ}{mol} } \)
\( \mathrm { \frac{1}{2}H_2(g) + \frac{1}{2}Cl_2(g) \rightarrow HCl(g) } \) \( \mathrm { \Delta_fH^{\circ} = -92.3 \frac{kJ}{mol} } \)


Next, predict the standard molar enthalpy of decomposition by reversing the reactions. Remember also to reverse the signs of the enthalpies.

\( \mathrm { HBr(g) \rightarrow \frac{1}{2}H_2(g) + \frac{1}{2}Br_2(l) } \) \( \mathrm { \Delta_dH^{\circ} = +36.3 \frac{kJ}{mol} } \)
\( \mathrm { HCl(g) \rightarrow \frac{1}{2}H_2(g) + \frac{1}{2}Cl_2(g) } \) \( \mathrm { \Delta_dH^{\circ} = +92.3 \frac{kJ}{mol} } \)


The simple decomposition of HBr requires 36.3 kJ/mol whereas the simple decomposition of HCl requires 92.3 kJ/mol. Because HCl requires more energy to decompose, HCl is more stable than HBr. Remember that the lower the standard molar enthalpy of formation, the higher the standard molar enthalpy of decomposition. The higher the standard molar enthalpy of decomposition, the greater the thermal stability!



Fig. 1


Note that some compounds have positive enthalpies of formation (Refer to the Table of Standard Molar Enthalpies of Formation in your data booklet). These compounds release heat when they decompose into their elements.

Read page 510 in the textbook.

Check Your Understanding


Complete Section 11.5 Question 1 on page 514 of the textbook. Click on the link below to check your work.

Page 514 Section 11.5 Question 1

ΔfH° for methanol = -239.2 kJ/mol

ΔfH° for ethanol = -277.6 kJ/mol

Since we know the enthalpies of formation of the two compounds (refer to the data booklet), we can predict their enthalpies of decomposition:

ΔdH° for methanol = +239.2 kJ/mol

ΔdH° for ethanol = +277.6 kJ/mol

Since it takes more energy to decompose ethanol back to its elements, therefore ethanol is more stable.



  1.  
  2. Predicting Standard Enthalpies of Reaction


    So far in this module you have learned two different methods of determining enthalpy change:

  • In lesson 1, you learned how to calculate experimental enthalpy change using calorimetric data.

  • In lesson 2, you manipulated sets of reactions to predict a theoretical enthalpy change for a reaction.


But what if you are asked to predict the enthalpy change of a reaction for which you have no calorimetric data and no set of reactions? How will you predict the enthalpy change? In this case, you may use an equation that utilizes the Table of Standard Molar Enthalpies of Formation in combination with Hess' Law. The equation is as follows:

\( \mathrm { \Delta_rH^{\circ} = \Sigma n\Delta_{fP}H_m^{\circ} - \Sigma n\Delta_{fR}H_m^{\circ} } \)


Basically, this equation states that the net enthalpy change for a chemical reaction is the total chemical potential energy of the products minus total chemical potential energy of the reactants.

Refer to Sample Problem 11.5 on page 511 of your textbook in order to better understand how this equation was derived.


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For more practice using enthalpies of formation, study the two additional examples provided below.

Example 1

Calculate the standard enthalpy of reaction for the roasting of zinc sulfide, as represented by the following reaction:

\( \mathrm { 2ZnS(s) + 3O_2(g) \rightarrow 2ZnO(s) + 2SO_2(g) } \)
\( \mathrm { \Delta_rH^{\circ} = ? } \)

Reactants Products
Species ZnS(s) O2(g) ZnO(s) SO2(g)
Coefficient (mol) 2 3 2 2
ΔfHm° (kJ/mol) -206.0 0 -350.5 -296.8
ΔrH° ΔrH° = ΣnΔfPHm° - ΣnΔfRHm°
ΣnΔfpHm° - ΣnΔfrHm° =[(2 mol x -350.5 kJ/mol) + (2 mol x -296.8 kJ/mol)] - [(2 mol x -206.6 kJ/mol) + (3 mol x 0 kJ/mol)]
ΔrH° -882.6 kJ


Example 2

Predict the ΔfH° for the following formation reaction:

\( \mathrm { N_2 + 3H_2 \rightarrow 2NH_3(g) } \)
\( \mathrm { \Delta_fH^{\circ} = ? } \)

Reactants Products
Species N2(g) H2(g) NH3(g)
Coefficient (mol) 1 3 2
ΔfHm° (kJ/mol) 0 0 -45.9
ΣnΔfHm° = [(1 mol x 0 kJ/mol) + ( 3 mol x 0 kJ/mol) = 0 kJ = (2 mol x -45.9 kJ/mol) = -91.8 kJ
ΣnΔfpHm° - ΣnΔfrHm° = -91.8 kJ - (0kJ)
ΔfH° = -91.8 kJ
Read pages 512 to 513 in the textbook.

Check Your Understanding


Complete Section 11.5 Question 2 (a, b) and 3 (a, b, c) on page 514 of the textbook. Click on the link below to check your work.

Page 514 Section 11.5 Question 2 (a, b)

\( \mathrm { \Delta_rH^{\circ} = \Sigma n\Delta_{fp}H_m^{\circ} - \Sigma n\Delta_{fr}H_m^{\circ} } \)



  1. Reactants Products
    Species CH4(g) H2O(g) CO(g) H2(g)
    Coefficient (mol) 1 1 1 3
    ΔfHm° (kJ/mol) -74.6 -241.8 -110.5 0
    ΣnΔfHm° = [(1 mol x-74.6kJ/mol) + ( 1 mol x -241.8 kJ/mol)] = -316.4 kJ = [(1 mol x -110.5 kJ/mol) + (3 mol x 0kJ/mol)] = -110.5 kJ
    ΣnΔfpHm° minus ΣnΔfrHm° = -110.5 kJ - (-316.4 kJ) = +205.9 kJ


  2. Reactants Products
    Species CO(g) H2O(g) CO2(g) H2(g)
    Coefficient (mol) 1 1 1 1
    ΔfHm° (kJ/mol) -110.5 -241.8 -393.5 0
    ΣnΔfHm° = [(1 mol x-110.5 kJ/mol) + ( 1 mol x -241.8 kJ/mol)] = -352.3 kJ = [(1 mol x -393.5 kJ/mol) + (1 mol x 0 kJ/mol)] = -393.5 kJ
    ΣnΔfpHm° minus ΣnΔfrHm° = -393.5 kJ - (-352.3 kJ) = -41.2 kJ


Page 514 Section 11.5 Question 3 (a, b, c)

\( \mathrm { \Delta_rH^{\circ} = \Sigma n\Delta_{fp}H_m^{\circ} - \Sigma n\Delta_{fr}H_m^{\circ} } \)



  1. Reactants Products
    Species NH3(g) O2(g) NO(g) H2O(g)
    Coefficient (mol) 4 5 4 6
    ΔfHm° (kJ/mol) -45.9 0 +91.3 -241.8
    ΣnΔfHm° = [(4 mol x -45.9 kJ/mol) + ( 5 mol x 0 kJ/mol) = -183.6 kJ = [(4 mol x +91.3 kJ/mol) + (6 mol x -241.8 kJ/mol)] = -1085.6 kJ
    ΣnΔfpHm° minus ΣnΔfrHm° = -1085.6 kJ - (-183.6 kJ) = -902.0 kJ


    \( \mathrm { \Delta_rH_m^{\circ} = \dfrac{-902.0~kJ}{4} = -225.5 \frac{kJ}{mol~of~ammonia} } \)



  2. Reactants Products
    Species NO(g) O2(g) NO2(g)
    Coefficient (mol) 2 1 2
    ΔfHm° (kJ/mol) +91.3 0 +33.2
    ΣnΔfHm° = [(2 mol x +91.3 kJ/mol) + ( 1 mol x 0 kJ/mol) = +182.6 kJ = (2 mol x +33.2 kJ/mol)
    ΣnΔfpHm° minus ΣnΔfrHm° = +66.4 kJ - (+182.6 kJ) = -116.2 kJ


    \( \mathrm { \Delta_rH_m^{\circ} = \dfrac{-116.2~kJ}{2} = -58.1 \frac{kJ}{mol~of~nitrogen~monoxide} } \)



  3. Reactants Products
    Species NO2(g) H2O(l) HNO3(l) NO(g)
    Coefficient (mol) 3 1 2 1
    ΔfHm° (kJ/mol) +33.2 -285.8 -174.1 +91.3
    ΣnΔfHm° = [(3 mol x +33.2kJ/mol) + ( 1 mol x -285.8 kJ/mol) = -186.2kJ = [(2 mol x -174.1 kJ/mol) + (1 mol x +91.3 kJ/mol)] = -256.9 kJ
    ΣnΔfpHm° minus ΣnΔfrHm° = -256.9 kJ - (-186.2 kJ) = -70.7 kJ


    \( \mathrm { \Delta_rH_m^{\circ} = \dfrac{-70.7~kJ}{3} = -23.6 \frac{kJ}{mol~of~nitrogen~dioxide} } \)





  1.  
  2.  
  3. Predicting Standard Molar Enthalpies of Formation


    In the previous Self-Check activity, you used the standard molar enthalpies of formation for compounds to predict the standard enthalpy of a reaction. Can the reverse be determined? That is, can the standard molar enthalpy of formation for a compound be predicted using the standard enthalpy of a reaction? The answer is yes - as long as the standard molar enthalpies of formation for the other reaction species are known. Study the example below:



Example

Calculate the standard molar enthalpy of formation for hexane, C6H14(l). Assume that its standard molar enthalpy of combustion is -4162.9 kJ/mol.


\( \mathrm { C_6H_{14}(l) + \frac{19}{2}O_2(g) \rightarrow 6CO_2(g) + 7H_2O(g) } \)


\( \mathrm { \Delta_rH_m^{\circ}~for~C_6H_{14}(l) = -4162.9 \frac{kJ}{mol} } \)


\( \mathrm { \Delta_rH^{\circ} = -4162.9 \frac{kJ}{mol} \times 1~mol~C_6H_{14}(l) = -4162.9 kJ } \)



Reactants Products ΔrH°
Species C6H14(l) O2(g) CO2(g) H2O(g)
Coefficient (mol) 1 19/2 6 7
ΔfHm° (kJ/mol) ? 0 -393.5 -241.8
ΣnΔfHm° = [(1 mol ΔfHm°(hexane)) + (0 kJ/mol)]

= 1 mol x ΔfHm°(hexane) 		= [(6 mol x -393.5 kJ/mol) + (7 mol x -241.8 kJ/mol)]

= -4053.6 kJ 		-4162.9 kJ

\( \mathrm { -4053.6 kJ - (1 mol \times \Delta_fH_{m~(hexane)}^{\circ}) = -4162.9 kJ } \)


\( \mathrm { \Delta_fH_{m~(hexane)}^{\circ} = +109.3 \frac{kJ}{mol} } \)


The standard molar enthalpy of formation for hexane is +109.3 kJ/mol.


Learning Tip

Take note of the state of H2O that is produced; the enthalpies of formation for liquid water and water vapour are different. In a closed system, such as a calorimeter, H2O(l) is formed. In an open system, H2O(g) is formed.


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