Write a balanced equation for the combustion of c7h16(l) (heptane) -- i. e. its reaction with o2(g) forming the products co2(g) and h2o(l). given the following standard heats of formation: δhf° of co2(g) is -393.5 kj/mol δhf° of h2o(l) is -286 kj/mol δhf° of c7h16(l) is -187.8 kj/mol what is the standard heat of reaction (δh°) for the combustion reaction of c7h16(l)? -4854.7 kj calculate the difference, δh-δe=δ(pv) for the combustion reaction of 1 mole of heptane. (assume standard state conditions and 298 k for all reactants and products.)

The standard enthalpy of reaction = -4854.7kJ

The difference: ΔH-ΔE = Δ(PV) = Δn.R.T = 9910.288 J ≈ 9.91 kJ    

Explanation:

The balanced chemical equation for the combustion of heptane:

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

To calculate the standard enthalpy of reaction () can be calculated by the Hess's law:

Here, is the stoichiometric coefficient

⇒

⇒

⇒

To calculate the difference: 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) = 9910.288 J = 9.91 kJ                              (∵ 1 kJ = 1000J )

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