Acetylene gas C?H? can be produced by the reaction of calcium carbide CaC? with water according to the following balanced equation: CaC?(s) + 2 H?O(l)? Ca(OH)?(s) + C?H?(g) Determine the value of ?rH° for this reaction using the thermochemical equations provided below: Ca(s) + 2 C(gr) ? CaC?(s) ?rH° = -62.8 kJ mol?¹ Ca(s) + ½ O?(g) ? CaO(s) ?rH° = -635.5 kJ mol?¹ CaO(s) + H?O(l) ? Ca(OH)?(s) ?rH° = -65.2 kJ mol?¹ 2 C?H?(g) + 5 O?(g) ? 4 CO?(g) + 2 H?O(l) ?rH° = -2600 kJ mol?¹ C(gr) + O?(g) ? CO?(g) ?rH° = -393.5 kJ mol?¹
The Correct Answer and Explanation is:
To determine the enthalpy change (ΔrH°) for the reaction:
CaC₂(s) + 2 H₂O(l) → Ca(OH)₂(s) + C₂H₂(g)
We will use Hess’s Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes of the steps into which the reaction can be divided.
Step 1: Write out the given thermochemical equations
- Ca(s) + 2 C(graphite) → CaC₂(s)
ΔH° = -62.8 kJ/mol - Ca(s) + ½ O₂(g) → CaO(s)
ΔH° = -635.5 kJ/mol - CaO(s) + H₂O(l) → Ca(OH)₂(s)
ΔH° = -65.2 kJ/mol - 2 C₂H₂(g) + 5 O₂(g) → 4 CO₂(g) + 2 H₂O(l)
ΔH° = -2600 kJ/mol
(Per mole of C₂H₂: divide by 2 → -1300 kJ/mol) - C(graphite) + O₂(g) → CO₂(g)
ΔH° = -393.5 kJ/mol
Step 2: Build a thermochemical cycle
We need to rearrange and combine the given reactions to reflect:
CaC₂(s) + 2 H₂O(l) → Ca(OH)₂(s) + C₂H₂(g)
a) Formation of CaC₂ (from Ca and C):
Already given:
Ca(s) + 2 C(gr) → CaC₂(s) → ΔH = -62.8 kJ
b) Convert Ca(s) into Ca(OH)₂:
Ca(s) + ½ O₂ → CaO(s) → ΔH = -635.5 kJ
CaO(s) + H₂O(l) → Ca(OH)₂(s) → ΔH = -65.2 kJ
Total for Ca(s) + H₂O → Ca(OH)₂(s):
ΔH = -635.5 + (-65.2) = -700.7 kJ
But we need Ca(OH)₂ on the product side, so we reverse this:
Ca(OH)₂(s) → Ca(s) + ½ O₂ + H₂O(l) → ΔH = +700.7 kJ
c) Combustion of C₂H₂:
C₂H₂(g) + 2.5 O₂ → 2 CO₂ + H₂O → ΔH = -1300 kJ
We reverse this to form C₂H₂:
2 CO₂ + H₂O → C₂H₂ + 2.5 O₂ → ΔH = +1300 kJ
d) CO₂ from graphite:
2 C(gr) + 2 O₂ → 2 CO₂ → ΔH = 2 × (-393.5) = -787.0 kJ
Step 3: Apply Hess’s Law
Use:
- Formation of CaC₂: -62.8 kJ
- Reverse of Ca(OH)₂ formation: +700.7 kJ
- Reverse of C₂H₂ combustion: +1300 kJ
- Formation of CO₂ from C: -787.0 kJ
Add all:
ΔrH° = -62.8 + 700.7 + 1300 + (-787.0)
ΔrH° = +1150.9 kJ
Final Answer:
ΔrH° = +1150.9 kJ/mol
Explanation
To calculate the enthalpy change (ΔrH°) for a chemical reaction that is not directly listed among standard thermochemical equations, we use Hess’s Law. This law states that the total enthalpy change for a reaction is independent of the path taken and depends only on the initial and final states.
In this case, we are determining the enthalpy change for the reaction of calcium carbide (CaC₂) with water to form calcium hydroxide (Ca(OH)₂) and acetylene gas (C₂H₂). The approach is to break this overall reaction into known steps for which enthalpy values are provided.
First, we used the formation of CaC₂ from calcium metal and graphite. Then, we considered the reverse of calcium hydroxide formation to break it down into its elemental components. We also reversed the combustion of acetylene to reflect its formation from CO₂ and H₂O. Finally, we included the formation of CO₂ from graphite to close the cycle.
By summing the enthalpies of these adjusted reactions, Hess’s Law allows us to derive the enthalpy change for the target reaction. The result is a positive enthalpy change of +1150.9 kJ/mol, meaning the reaction is endothermic. This indicates that the process requires a significant input of energy to proceed, which is consistent with the nature of reactions that produce gaseous hydrocarbons like acetylene.
