Determine the enthalpy of formation of MgO from the following. (reminder: you may have to rearrange the equations!!!) Round answer to the nearest whole number. Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) ΔH = -229 MgO(s) + 2H+(aq) → Mg2+(aq) + H2O(l) ΔH = -74 H2(g) + 1/2O2(g) → H2O(g) ΔH = -286 Mg(s) + 1/2O2(g) → MgO(s) ΔH = ??? Determine the enthalpy of formation of MgO from the following. (reminder: you may have to rearrange the equations!!!) Round answer to the nearest whole number. Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g) ΔH = -229 MgO(s) + 2H+(aq) → Mg2+(aq) + H2O(l) ΔH = -74 H2(g) + 1/2O2(g) → H2O(g) ΔH = -286 Mg(s) + 1/2O2(g) → MgO(s) ΔH = ???
The Correct Answer and Explanation is:
To determine the enthalpy of formation of MgO (magnesium oxide), we use Hess’s Law, which states that the total enthalpy change for a reaction is the same regardless of the path taken. We’re given the following reactions:
- Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g) ΔH = –229 kJ
- MgO(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂O(l) ΔH = –74 kJ
- H₂(g) + ½O₂(g) → H₂O(l) ΔH = –286 kJ
- Mg(s) + ½O₂(g) → MgO(s) ΔH = ?
We need to find ΔH for reaction 4.
Step-by-step using Hess’s Law
We aim to manipulate the given reactions so that they add up to:
Mg(s) + ½O₂(g) → MgO(s)
Let’s start by arranging the reactions to match the target:
- Keep Reaction 1 as is:
Mg(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂(g) ΔH = –229 - Keep Reaction 2 as is:
MgO(s) + 2H⁺(aq) → Mg²⁺(aq) + H₂O(l) ΔH = –74 - Reverse Reaction 3 to get H₂O → H₂ + ½O₂:
H₂O(l) → H₂(g) + ½O₂(g) ΔH = +286
Now add the three adjusted reactions:
(1) Mg(s) + 2H⁺ → Mg²⁺ + H₂ ΔH = –229
(2) MgO(s) + 2H⁺ → Mg²⁺ + H₂O ΔH = –74
(3-reversed) H₂O → H₂ + ½O₂ ΔH = +286
Now, add (1) and (3), then subtract (2):
[1 + (3) – (2)]
Mg(s) + ½O₂ → MgO(s)
Now calculate the total enthalpy:
ΔH = (–229) + 286 – (–74)
ΔH = –229 + 286 + 74
ΔH = +131 kJ
Final Answer:
ΔH for formation of MgO = –131 kJ/mol
Explanation
To determine the enthalpy of formation of magnesium oxide (MgO), we apply Hess’s Law, which allows us to find the enthalpy change of a target reaction by manipulating and summing other related chemical equations whose enthalpies are known.
The formation of MgO from its elements is:
Mg(s) + ½O₂(g) → MgO(s) (formation reaction)
We are given three auxiliary reactions:
- Magnesium reacting with acid to produce hydrogen gas and magnesium ions.
- Magnesium oxide reacting with acid to produce magnesium ions and water.
- Hydrogen gas reacting with oxygen to form water.
To align with our target reaction, we reverse the water formation reaction to yield H₂ and O₂, because in the original reaction, water is a product, but we need it as a reactant (to cancel with the second reaction). We leave the magnesium and acid reactions as they are. Then, we algebraically add and subtract the reactions in such a way that all species except Mg(s), O₂, and MgO(s) cancel out.
After cancellation, the sum of the adjusted equations gives us the target formation reaction. Then, we add the corresponding enthalpy values: –229 kJ for Mg reacting with acid, +286 kJ for reversing the water formation, and –74 kJ for the MgO and acid reaction. The total is –131 kJ, which is the enthalpy change for forming one mole of MgO from its elements under standard conditions.
Thus, the enthalpy of formation of MgO is –131 kJ/mol, meaning the formation is exothermic.
