Calculate the standard enthalpy of formation of solid magnesium hydroxide, given the following data:
2Mg(s) + O₂(g) → 2MgO(s) ∆H° = – 1203.6 kJ
Mg(OH)₂(s)→ MgO(s) + H₂O(1) ∆H° = +37.1 kJ
2H₂(g) + O₂(g) → 2H₂O(1) ∆H° = -571.7 kJ
The Correct Answer and Explanation is :
To calculate the standard enthalpy of formation of solid magnesium hydroxide (\text{Mg(OH)}_2), we can use Hess’s law. The standard enthalpy of formation (\Delta H_f^\circ) refers to the enthalpy change when 1 mole of a substance is formed from its elements in their standard states.
The reaction for the standard enthalpy of formation of (\text{Mg(OH)}_2) is:
[
\text{Mg}(s) + \frac{1}{2}\text{O}_2(g) + \text{H}_2(g) \rightarrow \text{Mg(OH)}_2(s)
]
We are given the following reactions:
- (2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)) (\Delta H^\circ = -1203.6\ \text{kJ}) For 1 mole of MgO:
[
\Delta H^\circ = \frac{-1203.6}{2} = -601.8\ \text{kJ/mol}
] - (\text{Mg(OH)}_2(s) \rightarrow \text{MgO}(s) + \text{H}_2O(l)) (\Delta H^\circ = +37.1\ \text{kJ/mol})
- (2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2O(l)) (\Delta H^\circ = -571.7\ \text{kJ}) For 1 mole of (\text{H}_2O):
[
\Delta H^\circ = \frac{-571.7}{2} = -285.85\ \text{kJ/mol}
]
Applying Hess’s Law:
The desired reaction is:
[
\text{Mg}(s) + \frac{1}{2}\text{O}_2(g) + \text{H}_2(g) \rightarrow \text{Mg(OH)}_2(s)
]
To construct this, we reverse reaction 2 to form (\text{Mg(OH)}_2(s)) from (\text{MgO}(s)) and (\text{H}_2O(l)), and we use reactions 1 and 3.
[
\text{MgO}(s) + \text{H}_2O(l) \rightarrow \text{Mg(OH)}_2(s) \ \ \ \Delta H^\circ = -37.1\ \text{kJ/mol}
]
Now combine the enthalpies:
[
\Delta H^\circ = (-601.8\ \text{kJ/mol}) + (-285.85\ \text{kJ/mol}) + (-37.1\ \text{kJ/mol})
]
[
\Delta H^\circ = -924.75\ \text{kJ/mol}
]
Thus, the standard enthalpy of formation of (\text{Mg(OH)}_2(s)) is -924.75 kJ/mol.
This result was derived by combining reactions to ensure all intermediate steps cancel out, leaving only the formation of magnesium hydroxide from its elemental components. Hess’s law allows us to add the enthalpy changes of the reactions because enthalpy is a state function, meaning its change depends only on the initial and final states, not on the path taken.