Question

25.

Calculate the standard enthalpy change for the following chemical equation.

4FeO (s) + O2 (g) → 2Fe2O3 (s)

Use the following thermochemical equations to solve for the change in enthalpy.

Fe (s) + ½ O2 (g) → FeO (s)   ΔH = -269 kJ/mol

2Fe (s) + 3/2 O2 (g) → Fe2O3 (s)                  ΔH = -825 kJ/mol

-2726 kJ/mol

556 kJ/mol

-556 kJ/mol

574 kJ/mol

-574 kJ/mol

Answer 1

4FeO (s) + O2 (g) → 2Fe2O3 (s)

Use the following thermochemical equations to solve for the change in enthalpy.

Fe (s) + ½ O2 (g) → FeO (s)   ΔH = -269 kJ/mol -------------(1)

2Fe (s) + 3/2 O2 (g) → Fe2O3 (s)                  ΔH = -825 kJ/mol ---------------(2)

Multiply equation (2) with 2

4Fe (s) + 3 O2 (g) → 2Fe2O3 (s)                  ΔH = -1650 kJ ---------------(3)

Multiply equation (1) with 4

4Fe (s) + 2O2 (g) →4 FeO (s)   ΔH = -1076 kJ -------------(4)

Subtract equation 4 from equation 3

4FeO (s) + O2 (g) → 2Fe2O3 (s) ΔH = -1650 kJ-(-1076 kJ) = -1650+1076 = -574 kJ

the standard enthalpy change for the above chemical equation is  -574 kJ/mol.