The standard enthalpy change for the given reactions are:
a. ΔH° = -484.4 kJ/mol.
b. ΔH° = 106.5 kJ/mol.
c. ΔH° = 1,146.2 kJ/mol.
To calculate the standard enthalpy change (ΔH°) for the given reactions, we need to use the standard enthalpies of formation table. We can calculate ΔH° by taking the sum of the products' enthalpies of formation multiplied by their coefficients and subtracting the sum of the reactants' enthalpies of formation multiplied by their coefficients.
a. Mg(OH)₂(s) + 2NH₂*(aq) → Mg¹²(aq) + 2NH₂(g) +H₂O)
The enthalpy of formation of Mg(OH)₂(s) is -924.7 kJ/mol.
The enthalpy of formation of Mg²⁺(aq) is 0 kJ/mol.
The enthalpy of formation of NH₂(g) is 111.7 kJ/mol.
The enthalpy of formation of NH₂*(aq) is -133.3 kJ/mol.
The enthalpy of formation of H₂O(l) is -285.8 kJ/mol.
Therefore, ΔH° = (0 + 2*111.7 - 133.3 - 285.8) - (-924.7) = -484.4 kJ/mol.
b. PbO(s) + C(s) → CO(g) + Pb(s)
The enthalpy of formation of PbO(s) is -217.0 kJ/mol.
The enthalpy of formation of C(s) is 0 kJ/mol.
The enthalpy of formation of CO(g) is -110.5 kJ/mol.
The enthalpy of formation of Pb(s) is 0 kJ/mol.
Therefore, ΔH° = (0 - 110.5) - (-217.0 + 0) = 106.5 kJ/mol.
c. Mn(s) + 4H*(aq) + SO₂²(aq) → Mn²(aq) + SO₂(g) + 2H₂O(1)
The enthalpy of formation of Mn(s) is 0 kJ/mol.
The enthalpy of formation of H*(aq) is 0 kJ/mol.
The enthalpy of formation of SO₂²(aq) is -296.8 kJ/mol.
The enthalpy of formation of Mn²(aq) is 0 kJ/molThe enthalpy of formation of SO₂(g) is -296.8 kJ/mol.
The enthalpy of formation of H₂O(l) is -285.8 kJ/mol.
Therefore, ΔH° = (0 + 40 - 296.8) - (0 + 0 - 296.8 - 2-285.8) = 1,146.2 kJ/mol.