Find the enthalpy of formation of acetylene (C2H2): 2C+H2 C2H2If the enthalpy of combustion of C, H2 and C2H2 are -400kj.mol-1, -300kj.mol-1 and-1300kj.mol-1?
Question
Find the enthalpy of formation of acetylene (C2H2): 2C+H2 C2H2If the enthalpy of combustion of C, H2 and C2H2 are -400kj.mol-1, -300kj.mol-1 and-1300kj.mol-1?
Solution
The enthalpy of formation of a compound is calculated using the formula:
ΔHf = Σ ΔHf (products) - Σ ΔHf (reactants)
In this case, the reaction is: 2C + H2 → C2H2
The enthalpy of formation of the reactants is given as:
ΔHf (C) = -400 kJ/mol ΔHf (H2) = -300 kJ/mol
The enthalpy of formation of the product (C2H2) is what we're trying to find.
We can rearrange the formula to solve for ΔHf (C2H2):
ΔHf (C2H2) = Σ ΔHf (reactants) - Σ ΔHf (products)
Substituting the given values:
ΔHf (C2H2) = [2*(-400 kJ/mol) + (-300 kJ/mol)] - (-1300 kJ/mol)
ΔHf (C2H2) = -1100 kJ/mol - (-1300 kJ/mol)
ΔHf (C2H2) = 200 kJ/mol
So, the enthalpy of formation of acetylene (C2H2) is 200 kJ/mol.
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