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Enthalpy change of Pentane (C5H12) that burns is -3,270.5 kJ
Enthalpy change is defined as the difference in enthalpy of total products and reactants, each multiplied by the respective moles. It is expressed as ΔH°rxn
The formula for calculating the enthalpy change of a reaction is:
ΔH°rxn =∑[n x ΔH°f(product) - ∑[n x ΔH°f(reactant)]
The equilibrium reactions are:
C₅H₁₂(l) + 8O₂(g) ⇒ 5CO₂(g) +6H₂O(g)
ΔH°f(CO₂)(g) = -393.5 kJ/mol
ΔH°f(H₂O)(g) = - 241.8 kJ/mol
ΔH°f(C₅H₁₂)(l) = -146.8 kJ/mol
ΔH°f(O₂)(g) = 0 kJ/mol
The equation for the enthalpy change for the above reaction is:
ΔH°rxn =∑[n x ΔH°f(product) - ∑[n x ΔH°f(reactant)]
ΔH°rxn =∑[(n(CO₂) x ΔH°f(CO₂)) + (n(H₂O) x ΔH°f(H₂O)) ]- ∑[(n x ΔH°f(C₅H₁₂)) + (n x ΔH°f(O₂))]
ΔH°rxn =[( 5 x -393.5) + (6 x - 241.8 ) ]- [(1 x -147.8) + (8 x O₂)]
ΔH°rxn = -3,418.3 -(-147.8)
ΔH°rxn = -3,270.5 kJ
Therefore, the enthalpy of burning 1 mole of pentane is -3,270.5 kJ
learn more about enthalpy at https://brainly.com/question/14048672
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