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Question: Calculate the enthalpy of formation of ethyl alcohol from the following data: \({C_2}{H_5}OH(l) +...

Calculate the enthalpy of formation of ethyl alcohol from the following data:
C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l);ΔrHo=−1368.0kJ{C_2}{H_5}OH(l) + 3{O_2}(g) \to 2C{O_2}(g) + 3{H_2}O(l);{\Delta _r}{H^o} = - 1368.0kJ
C(s)+O2(g)→CO2(g);ΔrHo=−393.5kJC(s) + {O_2}(g) \to C{O_2}(g);{\Delta _r}{H^o} = - 393.5kJ
H2(g)+12O2(g)→H2O(l);ΔrHo=−286.0kJ{H_2}(g) + \dfrac{1}{2}{O_2}(g) \to {H_2}O(l);{\Delta _r}{H^o} = - 286.0kJ

Explanation

Solution

The change in enthalpy occurring during the formation of one mole of a substance from its constituent elements is termed as the standard enthalpy of formation of the compound. The heat released on combustion of one mole of a substance is known as the enthalpy of combustion of the substance.
Formula used : ΔrHo=ΔfHo(products)−ΔfHo(reactants){\Delta _r}{H^o} = {\Delta _f}{H^o}\left( {{\text{products}}} \right) - {\Delta _f}{H^o}\left( {{\text{reactants}}} \right)

Complete step by step answer:
In the question, we are given the enthalpy of combustion of ethyl alcohol=−1368.0kJ = - 1368.0kJ
The enthalpy of formation of carbon dioxide =−393.5kJ = - 393.5kJ
The enthalpy of formation of water =−286.0kJ = - 286.0kJ
We know that the enthalpy of a reaction is given by the difference between the enthalpies of formation of the reactants and the enthalpies of formation of the reactants.
Or we can write it as, ΔrHo=ΔfHo(products)−ΔfHo(reactants){\Delta _r}{H^o} = {\Delta _f}{H^o}\left( {{\text{products}}} \right) - {\Delta _f}{H^o}\left( {{\text{reactants}}} \right)
For the reaction, C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l);[ΔrHo=−1368.0kJ]{C_2}{H_5}OH(l) + 3{O_2}(g) \to 2C{O_2}(g) + 3{H_2}O(l);\left[ {{\Delta _r}{H^o} = - 1368.0kJ} \right]
We can write the equation as,
ΔrHo=(2×ΔfH(CO2)o)+(3×ΔfH(H2O)o)−(ΔfH(C2H5OH)o){\Delta _r}{H^o} = \left( {2 \times {\Delta _f}H_{\left( {C{O_2}} \right)}^o} \right) + \left( {3 \times {\Delta _f}H_{\left( {{H_2}O} \right)}^o} \right) - \left( {{\Delta _f}H_{\left( {{C_2}{H_5}OH} \right)}^o} \right)
Substituting the values into the formula, we get,
−1368=(2×(−393.5))+(3×(−286))−(ΔfH(C2H5OH)o)- 1368 = \left( {2 \times \left( { - 393.5} \right)} \right) + \left( {3 \times \left( { - 286} \right)} \right) - \left( {{\Delta _f}H_{\left( {{C_2}{H_5}OH} \right)}^o} \right)
We get, (ΔfH(C2H5OH)o)=−277kJ\left( {{\Delta _f}H_{\left( {{C_2}{H_5}OH} \right)}^o} \right) = - 277kJ
Thus, the enthalpy of formation of ethyl alcohol is −277kJ - 277kJ.

Note:
-The enthalpy of formation of a substance in its standard state is taken as zero. In the above reaction, we take the value of enthalpy of formation of oxygen as zero as it is present in its standard state, that is, the gaseous state.
The standard state is generally is that state:
-For a gas, which obeys the ideal gas equation at pressure of 1 bar
-For a solute, which is present in an ideal solution of concentration of unity
-For an element, which is the most stable in standard conditions.
-If the enthalpy of formation of a substance is negative, it is considered to be stable.