Chapter 10: Thermochemistry

Question 1

Use Hess’s Law to calculate the enthalpy change delta H (in kj) for the following reaction
NO2(g) + CO(g) -> NO(g) + CO2
given the following thermochemical equations:
2NO(g) + O2(g) -> 2 NO2 (g) H= -11.3.1 kj
2 CO(g) + O2(g) -> 2 CO2(g) H= -566.0 kj

Answer 1

Flip reaction 1, keep reaction 2 the same
Add the reactions together, cancel O2. Add the delta H values together
Divide the reaction and delta H by 2
NO2(g) + CO(g) -> NO(g) + CO2(g) delta H = -226.5 kj

Question 2

When magnesium metal is burned in a pure carbon dioxide atmosphere, carbon and magnesium oxide form. the balanced equation is
2Mg(s) + CO2(g) -> 2MgO(s) + C(s) delta Hrxn
Delta H for CO2 = -393.5 kJ/mole
Delta H for MgO = -601.6 kJ/mole
Use these enthalpies of formation (delta H) to calculate delta Hrxn

Answer 2

delta Hrxn= Delta H products-delta H reactants
Delta Hrxn = (2 X MgO + C) - (2 X Mg + CO2)
delta Hrxn= (2 X -6.016 kj/mol + 0 kj/mole) - (2 X 0 kj/mole + -393.5 kj/mole)
Delta Hrxn = -809.7 kj/mole

Question 3

Calculate enthalpy of combustion of ethane using the average bond enthalpy values below
2CH3CH3(g) + 7O2(g) -> 4CO2(g) + 6H2O(l)
C-H = 413
C-C = 347
O=O = 498
C=O = 799
O-H = 463
What does the delta H value tell you about the stability of the reactant bonds compare to the products?

Answer 3

delta H = (bonds broken) - (bonds formed)
delta H = (12CH + 2CC + 7O2) - (8CO + 12OH)
delta H = (12 X 413 + 2 X 347 + 7 X 498) - (8 X 799 + 12 X 463)
delta H = -2,812 kJ
since delta H is exothermic, this indicated that the bonds in the products are more stable than the bonds in the reactants. Energy was released as heat when the new bonds formed