Answer:
<u>The standard enthalpy of reaction = -4854.7kJ</u>
<u>The difference: </u>ΔH-ΔE = Δ(PV) = Δn.R.T = <u>9910.288 J ≈ 9.91 kJ</u>
Explanation:
<u>The balanced chemical equation for the combustion of heptane</u>:
C₇H₁₆ (l) + 11 O₂ (g) → 7 CO₂ (g) + 8 H₂O (l)
Given: The standard enthalpy of formation () for: C₇H₁₆ (l) = -187.8 kJ/mol, O₂ (g) = 0 kJ/mol, CO₂ (g) = -393.5 kJ/mol, H₂O (l) = -286 kJ/mol
<u>To calculate the standard enthalpy of reaction () can be calculated by the Hess's law</u>:
Here, is the stoichiometric coefficient
⇒
⇒
⇒
<u>To calculate the difference: </u>ΔH-ΔE=Δ(PV)
We use the ideal gas equation: P.V = n.R.T
⇒ ΔH-ΔE=Δ(PV) = Δn.R.T
Given: Temperature:T = 298K, R = 8.314 J⋅K⁻¹⋅mol⁻¹
Δn = number of moles of gaseous products - number of moles of gaseous reactants = (7)- (11) = (-4)
⇒ ΔH-ΔE=Δ(PV) = Δn.R.T = (-4 mol) × (8.314 J⋅K⁻¹⋅mol⁻¹) × (298K) = <u>9910.288 J = 9.91 kJ</u> (∵ 1 kJ = 1000J )
