Respuesta :
Answer:
226.2 kJ/mol
Explanation:
Let's consider the following thermochemical equation for the combustion of acetylene.
C₂H₂(g) + (5/2) O₂(g) → 2 CO₂(g) + H₂O(l), ΔH°rxn = –1299 kJ/mol.
We can also calculate the enthalpy of the reaction per mole of acetylene from the enthalpies of formation.
ΔH°rxn = 2 mol × ΔH°f(CO₂(g)) + 1 mol × ΔH°f(H₂O(l)) - 1 mol × ΔH°f(C₂H₂(g)) - 1 mol × ΔH°f(O₂(g))
1 mol × ΔH°f(C₂H₂(g)) = 2 mol × ΔH°f(CO₂(g)) + 1 mol × ΔH°f(H₂O(l)) - ΔH°rxn - 1 mol × ΔH°f(O₂(g))
1 mol × ΔH°f(C₂H₂(g)) = 2 mol × (-393.5 kJ/mol) + 1 mol × (-285.8 kJ/mol) - (-1299 kJ) - 1 mol × (0 kJ/mol)
ΔH°f(C₂H₂(g)) = 226.2 kJ/mol
Answer:
The enthalpy of formation of acetylene is 226.2 kJ/mol
Explanation:
Step 1: Data given
C2H2 (g) + (5/2)O2 (g) → 2CO2 (g) + H2O (l) Heat of Reaction (Rxn) = -1299kJ/mol
Standard formation [CO2 (g)]= -393.5 kJ/mol
Standard formation [H2O (l)] = -285.8 kj/mol
Step 2: The balanced equation
The formation of acetylene is:
2C(s) + H2(g) → C2H2(g)
Step 3: Calculate the enthalpy of formation of acetylene
It doesn't matter if the process will happen in 1 step or in more steps. What matters is the final result. This is Hess' law of heat summation.
To have the reaction of the formation of acetylene we have to take:
⇒ the reverse equation of the combustion of acetylene
2CO2 (g) + H2O (l) → C2H2 (g) + (5/2)O2 (g)
⇒ The equation of formation of CO2 (multiplied by 2)
2C(s) + 2O2(g) → 2CO2(g)
⇒ the equation of formation of H2O
H2(g) + 1/2 O2(g) → H2O(l)
2CO2 (g) + H2O (l) + 2C(s) + 2O2(g) + H2(g) + 1/2 O2(g → C2H2 (g) + (5/2)O2 (g) + 2CO2(g) + H2O(l)
Final reaction = 2C(s) + H2(g) → C2H2(g)
Calculate the enthalpy of formation of acetylene =
ΔHf = 1299 kJ/mol + (2*-393.5) kJ/mol + (-285.8) kJ/mol
ΔHf = 226.2 kJ/mol
The enthalpy of formation of acetylene is 226.2 kJ/mol
