Answer to Question #286914 in General Chemistry for yani

Question

the standard enthalpy of formation of liquid ethanol C2H5OH(I), is -227.63 kJ/mol. Calculate the heat of combustion at constant pressure and at constant volume of C2H5OH(I). ΔH°f, CO2= -393.5 kJ/mol; ΔH°f, H2O= -285.4 kJ/mol

C2H5OH(I)+3 O2(g) = 2 CO2(g) + 3 H2O

Answer 1

ΔH^°_f, C₂H₅OH(I) = −227.63 kJ/mol

ΔH^°_f, O₂(g) = 0 (simple substance)

ΔH^°_f, CO₂(g) = −393.5 kJ/mol

ΔH^°_f, H₂O(l) = −285.4 kJ/mol

Solution:

Balanced chemical equation:

C₂H₅OH(I) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

To find the heat (ΔH^°_r) of the ethanol combustion reaction, use the formula for the standard enthalpy change of formation:

ΔH^°_r = ∑ΔH^°_f (products) − ∑ΔH^°_f (reactants)

ΔH^°_r = 3 × ΔH^°_f, H₂O(l) + 2 × ΔH^°_f, CO₂(g) − 3 × ΔH^°_f, O₂(g) − ΔH^°_f, C₂H₅OH(I)

ΔH^°_r = 3 × (−285.4 kJ/mol) + 2 × (−393.5 kJ/mol) − 3 × 0 − (−227.63 kJ/mol) = −1415.57 kJ/mol

ΔH^°_r = −1415.57 kJ/mol

Answer: −1415.57 kJ/mol