Calculate and show the soluton of the standard enthalpy of formation of acetylene (C2H2) from its elements: 2C(graphite) + H2(g) →C2H2(g)
The equations for each step and the corresponding enthalpy changes are
C(graphite) + O2 → CO2(g) ∆HO
rxn = -393.5 kJ/mol
H2(g) + 1⁄2 O2(g) → H2O(l) ∆HO
rxn = -285.8 kJ/mol
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) ∆HO
rxn = -285.8 kJ/mol
The equations for each step and the corresponding enthalpy changes are
C(graphite) + O2 → CO2(g) ∆HO
rxn = -393.5 kJ/mol
H2(g) + 1⁄2 O2(g) → H2O(l) ∆HO
rxn = -285.8 kJ/mol
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) ∆HO
rxn = -285.8 kJ/mol
Answers & Comments
Answer:
Given reactions:
C(graphite) + O2 → CO2(g) ΔH°rxn = -393.5 kJ/mol
H2(g) + 1⁄2O2(g) → H2O(l) ΔH°rxn = -285.8 kJ/mol
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l) ΔH°rxn = -1300.6 kJ/mol
To find ΔH°f for:
2C(graphite) + H2(g) → C2H2(g)
Step 1. Reverse reaction 1:
CO2(g) → C(graphite) + O2 ΔH°rxn = +393.5 kJ/mol
Step 2. Reverse reaction 2:
H2O(l) → H2(g) + 1⁄2O2(g) ΔH°rxn = +285.8 kJ/mol
Step 3. Add the reversed reactions:
CO2(g) + H2O(l) → C(graphite) + H2(g) + 3/2O2(g) ΔH°rxn = +679.3 kJ/mol
Step 4. Multiply by 2 (for balancing):
2CO2(g) + 2H2O(l) → 2C(graphite) + 2H2(g) + 3O2(g) ΔH°rxn = +1358.6 kJ/mol
Step 5. Subtract reaction 3 from step 4:
2C(graphite) + 2H2(g) → C2H2(g)
ΔH°rxn = +1358.6 - (-1300.6) = +265.8 kJ/mol
Therefore, the standard enthalpy of formation (ΔH°f) of C2H2(g) is +265.8 kJ/mol.
Explanation: