5.6 Hess s Law

1. 5.6 Hess’s Law By Rebeka Cohan and Dory Glauberman

2. Definition Hess’s Law states that If a reaction is carried out in a series of steps, ?H for the reaction is the sum of ?H for each of the steps. The total change in enthalpy (?H) is independent of the number of steps and the order of the reactions.

3. Use Hess’s Law allows for the calculation of enthalpy data for reactions that are not easy to carry out directly. For example, C(s) + O2(g) produces a mixture of CO(g) and CO2(g) It provides an alternative way to calculate the change in enthalpy of a reaction. As opposed to making calorimetric measurements, you can find the change in enthalpy of one reaction by using tabulated ?H values of other reactions.

4. Example CH4(g) + 2O2(g) ? CO2(g) + 2H2O(g) ?H= -802 kJ 2H2O(g) ? 2H20(l) ?H= -88 kJ _______________________________________________ CH4(g) + 2O2(g) ? CO2(g) + 2H2O(l) ?H= -890 kJ Therefore for the reaction CH4(g) + 2O2(g) ? CO2(g) + 2H2O(l), the change in enthalpy (?H) is equal to -890 kJ

5. Note: ?H is dependent upon the amount of matter that undergoes the reaction, as well as the states of the reactants and products.

6. 5.61 Calculate the enthalpy change for the reaction P4O6(s) + 2O2(g) ? P4O10(s) Given the following enthalpies of reaction: P4(s) + 3O2(g) ? P4O6(s) ?H = -1640.1 kJ P4(s) + 5O2(g) ? P4O10(s) ?H = -2940.1 kJ

7. Solve P4(s) + 3O2(g) ? P4O6(s) ?H = -1640.1 kJ Transform the equation so that P4O6(s) is on the reactant side. P4O6(s) ? P4(s) + 3O2(g) -?H = 1640.1 kJ Note: if the equation switches, the sign of ?H does as well. P4(s) + 5O2(g) ? P4O10(s) ?H = -2940.1 kJ The second equation can stay the same because P4O10(s) is on the product side just like in the final equation.

8. Solve P4O6(s) ? P4(s) + 3O2(g) -?H = 1640.1 kJ P4(s) + 5O2(g) ? P4O10(s) ?H = -2940.1 kJ ____________________________________________________ P4O6(s) + 2O2(g) ? P4O10(s) ?H = -1300.0 kJ By adding the ?H for the two equations that add up to the desired reaction, you can find the ?H for that reaction.

9. 5.63 From the enthalpies of reaction H2(g) +F2(g) ? 2HF(g) ?H = -537 kJ C(s) + 2F2(g) ? CF4(g) ?H = -680 kJ 2C(s) + 2H2(g) ? C2H4(g) ?H = +52.3 kJ Calculate ?H for the reaction of ethylene with F2: C2H4(g) + 6F2(g) ? 2CF4(g) + 4HF(g)

10. Solve In order to keep balance with the final equation you need to multiply equations 1 and 2 by 2. 2(H2(g) +F2(g) ? 2HF(g) ?H = -537 kJ) 2(C(s) + 2F2(g) ? CF4(g) ?H = -680 kJ) 2H2(g) + 2F2(g) ? 4HF(g) ?H = -1074 kJ 2C(s) + 4F2(g) ?2 CF4(g) ?H = -1360 kJ The 3rd equations stays the same because it agrees with the final equation.

11. Solve 2H2(g) + 2F2(g) ? 4HF(g) ?H = -1074 kJ 2C(s) + 4F2(g) ?2 CF4(g) ?H = -1360 kJ 2C(s) + 2H2(g) ? C2H4(g) ?H = +52.3 kJ _____________________________________________________ C2H4(g) + 6F2(g) ? 2CF4(g) + 4HF(g) ?H = -2.49 X 103

12. More Problems 1. Calculate the standard enthalpy change, ?Ho, for the formation of 1 mol of strontium carbonate (the material that gives the red color in fireworks) from its elements.