Acetylene burns in air by the following equation:
C₂H₂(g) + 5/2 O₂ (g) → 2CO₂(g) + H₂O(g) ΔH°rxn = -1255.8kJ
Given ΔH°f of CO₂(g) = -393.5kJ/mol and ΔH°f of H₂O (g) = -241.8 k/mol, fimd ΔH°f of C₂H₂(g)

Respuesta :

The standard enthalpy of production of acetylene is 227.0 kJ/mol when acetylene combustion is taken into account.

The enthalpy change that takes place when 1 mole of a substance is created from its component elements in their standard states is known as the standard enthalpy of formation. A pure element has a standard enthalpy of production of 0 in its normal state.

Let's consider the following thermochemical equation.

C₂H₂(g) + 5/2 O₂(g) → 2 CO₂(g) + H₂O(g)    ΔH°rxn = −1255.8 kJ

We may determine the standard enthalpy of the reaction (H°rxn) using the following expression given the standard enthalpies of formation (H°f) and assuming that the enthalpy of formation of simple compounds in the standard state is zero.

ΔH°[tex]_{rxn}[/tex]= ∑ ΔH°[tex]_{f}[/tex](p)×[tex]n_{p}[/tex] ₋∑ ΔH°[tex]_{f}[/tex](r)×[tex]n_{r}[/tex]

where,

n: moles

r: reactants

p: products

This equation will be used to determine the standard enthalpy of acetylene production.

ΔH°[tex]_{rxn}[/tex]= ∑ ΔH°[tex]_{f}[/tex](p)×[tex]n_{p}[/tex] ₋∑ ΔH°[tex]_{f}[/tex](r)×[tex]n_{r}[/tex]

ΔH°[tex]_{f}[/tex](C₂H₂)(g) × 1mol = ΔH°[tex]_{f}[/tex](CO₂(g)) × 2mol + ΔH°[tex]_{f}[/tex](H₂O(g)) - ΔH°[tex]_{rxn}[/tex] - ΔH°[tex]_{f}[/tex]( O₂ (g)) ×5/2 mol

ΔH°[tex]_{f}[/tex](C₂H₂)(g) = 227.0kJ/mol

The standard enthalpy of production of acetylene is 227.0 kJ/mol

To know more about standard enthalpy refer to:  https://brainly.com/question/11417334

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