CHM 2045C Learning Tools

  Applications of Hess's Law

Hess's law states that if a reaction can be considered as the sum of a series of steps, than the enthalpy change for the reaction will be equal to the sum of the enthalpy changes for each of the steps.
You will need to refer to the Thermodynamic Table in your text to find the of reactants and products.

Sample Problem Set I

1. Write the equation for the combustion of propane gas (C₃H₈) under standard state conditions. Standard state is the normal physical state of the element or compound at 1 atm pressure and 25. Use the enthalpies of formation to determine the enthalpy of combustion of propane.

2. Write the equation for the enthalpy of formation of carbon dioxide gas and determine the enthalpy change for this reaction.

3. Write the equation for the enthalpy of formation of liquid water and determine the enthalpy change for this reaction.

4. Use the equations and enthalpy changes from 1,2 and 3 to determine the enthalpy change for the reaction.

5. Calculate the energy evolved by the reaction of 10.0 g of hydrogen with an excess of carbon.

Sample Problem Set II

1. Calculate the enthalpy change for the reaction: C₂H₄(g) + 6F₂(g) ® 2CF₄(g) + 4HF(g) using the following enthalpies of reaction: Don't forget to show how the reactions add up!
 
 

C(s) + 2F2(g) ® CF4(g) 2HF(g)  ® H2(g) + F2(g) 2C(s) + 2H2(g) ® C2H4(g)

H = -680 kJ  H = 537 kJ  H = 52.3 kJ

2. Using the answer from question 1, calculate the enthalpy change associated with the reaction of 55.39 g of fluorine gas with 7.21 g of C₂H₄ gas. Does this look like a limiting reagent problem to you? It should!

Answers

Sample Problem Set I

1. C₃H₈(g) + 5O₂(g) ® 3CO₂(g) + 4H₂O(l)

H = 3 mol CO₂(-393.509 kJ/mol) + 4 mol H₂O(-285.830) - 1 mol C₃H₈ (-103.8) = -2220.0 kJ

2. C(s) + O₂(g) ® CO₂(g) = -393.509 kJ/mol so H = -393.509 kJ

3. H₂(g) + 1/2 O₂(g) ® H₂O(l)= -285.830 kJ/mol so H = -285.830 kJ

4. Show how equations can be used as steps to add up to 3C(s) + 4H₂(g) ® C₃H₈(g). Cancel like terms.
 

3CO2(g) + 4H2O(l) ® C3H8(g) + 5O2(g)  3C(s) + 3O2(g) ® 3CO2(g)  4H2(g) + 2 O2 (g) ® 4H2O(l)	H = -1(-2220.0 kJ ) = 2220.0 kJ  H = 3(-393.509 kJ) = -1180.527 kJ  H = 4(-285.830 kJ ) = -1143.32 kJ
3C(s) + 4H2(g) ® C3H8(g)	H = 2220.0+(-1180.527)+(-1143.32) = -103.8 kJ

Compare the result with the for propane. They are the same. Why?
5. 103.8 kJ are evolved for every 4 moles of hydrogen that react so

10.0 g H₂ x (1mole H₂ / 2.016 gH₂) x (-103.8 kJ / 4 mole H₂) = -129 kJ

Problem Set II

1. We are aiming for C₂H₄(g) + 6F₂(g) ® 2CF₄(g) + 4HF(g) so
 

2C(s) + 4F2(g) ® 2CF4(g) 2H2(g) + 2F2(g) ® 4HF(g) C2H4(g) ® 2C(s) + 2H2(g)	H = 2(-680 kJ) = -1360 kJ  H = -2(537 kJ) = -1074 kJ  H = -1(52.3 kJ) = -52.3 kJ
C2H4(g) + 6F2(g) ® 2CF4(g) + 4HF(g)	H = -2486 kJ

2. Is F₂ or C₂H₄ the limiting reagent?

55.39 g F₂ x (1mol F₂/37.996 g F₂) = 1.457785 mol F₂

7.21 g C₂H₄ x (1 mol C₂H₄ / 28.053) = .2570135 mol C₂H₄

Compare the ratio with the equation: (1.457785 F₂/.2570135 C₂H₄) = 5.67 F₂/1C₂H₄ versus the equations ratio of 6F₂/1C₂H₄ so the F₂ is the limiting reagent!

-2486 kJ of heat are evolved for 6 moles of F₂ reacted so

1.457785 mol F₂ x (-2486 kJ)/( 6 mol F₂) = -604 kJ
Therefore 604 kJ of energy are evolved by the reaction of 55.39 g F₂ with an excess of C₂H₄.

Last modified March 9, 1999              Top        For more information, contact:  ksanchez@fccj.org