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