Question

Using the standard enthalpies of formation found in the textbook, determine the enthalpy change for the combustion of ethanol c2h5oh as given below. c2h5oh (l) + 3 o2(g) → 2 co2(g) + 3 h2o(g)

Answer 1

Enthalpy of formation is calculated by subtracting the total enthalpy of formation of the reactants from those of the products. This is called the HESS' LAW.
ΔHrxn = ΔH(products) - ΔH(reactants)

Since the enthalpies are not listed in this item, from reliable sources, the obtained enthalpies of formation are written below.
ΔH(C2H5OH) = -276 kJ/mol
ΔH(O2) = 0 (because O2 is a pure substance)
ΔH(CO2) = -393.5 kJ/mol
ΔH(H2O) = -285.5 kJ/mol

Using the equation above,
ΔHrxn = (2)(-393.5 kJ/mol) + (3)(-285.5 kJ/mol) - (-276 kJ/mol)
ΔHrxn = -1367.5 kJ/mol

Answer: -1367.5 kJ/mol

Answer 2

Answer: -1367.5 kJ

Explanation:

The balanced chemical reaction is,

$C_2H_5OH(l)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)$

The expression for enthalpy change is,

$\Delta H=\sum [n\times H_f(product)]-\sum [n\times H_f(reactant)]$

$\Delta H=[(n_{CO_2}\times H_f_{CO_2})+(n_{H_2O}\times H_f_{H_2O}) ]-[(n_{O_2}\times H_f_{O_2})+(n_{C_2H_5OH}\times H_f_{C_2H_5OH})]$

where,

n = number of moles

Now put all the given values in this expression, we get

$\Delta H=[(2\times -393.5 kJ/mol)+(3\times -285.5 k) ]-[(3\times 0)+(1\times -276]$

$\Delta H=-1367.5kJ$

Therefore, the enthalpy change for this reaction is, -1367.5 kJ