Use ∆Hf values (table found in notes) to solve the following problems. Determine the ∆H for each of the following reactions. Classify each reaction as either exothermic or endothermic.
a. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
∆H = ? kJ/mol
b. H2 (g) + Cl2(g) → 2HCl(g)
∆H = ? kJ/mol
c. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
∆H = ? kJ/mol
d. 2F2(g) + 2H2O(l) → 4HF(g) + O2(g) ∆H = ? kJ/mol
e. Na2O(s) + SO2(g) → Na2SO3(s)
∆H = ? kJ/mol
a. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
∆H = ? kJ/mol
b. H2 (g) + Cl2(g) → 2HCl(g)
∆H = ? kJ/mol
c. CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
∆H = ? kJ/mol
d. 2F2(g) + 2H2O(l) → 4HF(g) + O2(g) ∆H = ? kJ/mol
e. Na2O(s) + SO2(g) → Na2SO3(s)
∆H = ? kJ/mol
Answers & Comments
Verified answer
Answer:
To calculate the enthalpy change (∆H) for each reaction, we need to use the standard enthalpy of formation (∆Hf) values of each compound involved in the reaction. Here are the steps to solve each problem:
a. 3C(s) + Fe2O3(s) → 3CO(g) + 2Fe(s)
∆H = ? kJ/mol
Reactants:
- 3 moles of C(s): ∆Hf = 0 kJ/mol
- 1 mole of Fe2O3(s): ∆Hf = -824.2 kJ/mol
Products:
- 3 moles of CO(g): ∆Hf = -110.5 kJ/mol
- 2 moles of Fe(s): ∆Hf = 0 kJ/mol
∆H = [3(-110.5 kJ/mol) + 2(0 kJ/mol)] - [3(0 kJ/mol) + 1(-824.2 kJ/mol)]
∆H = -392.3 kJ/mol
b. 3Mn(s) + Fe2O3(s) → 3MnO(s) + 2Fe(s)
∆H = ? kJ/mol
Reactants:
- 3 moles of Mn(s): ∆Hf = 0 kJ/mol
- 1 mole of Fe2O3(s): ∆Hf = -824.2 kJ/mol
Products:
- 3 moles of MnO(s): ∆Hf = -385.2 kJ/mol
- 2 moles of Fe(s): ∆Hf = 0 kJ/mol
∆H = [3(-385.2 kJ/mol) + 2(0 kJ/mol)] - [3(0 kJ/mol) + 1(-824.2 kJ/mol)]
∆H = 289.2 kJ/mol
c. 12P(s) + 10Fe2O3(s) → 3P4O10(s) + 20Fe(s)
∆H = ? kJ/mol
Reactants:
- 12 moles of P(s): ∆Hf = 0 kJ/mol
- 10 moles of Fe2O3(s): ∆Hf = -824.2 kJ/mol
Products:
- 3 moles of P4O10(s): ∆Hf = -2980 kJ/mol
- 20 moles of Fe(s): ∆Hf = 0 kJ/mol
∆H = [3(-2980 kJ/mol) + 20(0 kJ/mol)] - [12(0 kJ/mol) + 10(-824.2 kJ/mol)]
∆H = 4708.6 kJ/mol
d. 3Si(s) + 2Fe2O3(s) → 3SiO2(s) + 4Fe(s)
∆H = ? kJ/mol
Reactants:
- 3 moles of Si(s): ∆Hf = 0 kJ/mol
- 2 moles of Fe2O3(s): ∆Hf = -824.2 kJ/mol
Products:
- 3 moles of SiO2(s): ∆Hf = -910.9 kJ/mol
- 4 moles of Fe(s): ∆Hf = 0 kJ/mol
∆H = [3(-910.9 kJ/mol) + 4(0 kJ/mol)] - [3