Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.
Sagot :
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
#SPJ4
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
