Question

Ethanol has been proposed as an alternative fuel. Calculate the standard enthalpy of combustion per gram of liquid ethanol.

Short answer

The standard enthalpy combustion of ethanol is -29.71 kJ/g.

Step-by-step solution

Enthalpy of formation of substance

The chemical reaction for combustion of ethanol is,

$C_2{H}_{5}OH\left(l\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+3H_{2}O\left(l\right)$

The enthalpy of formation of the substance present in the reaction is,

$$\begin{gathered} \Delta H_f^0 of C{O}_2\left(g\right)=-393.5kJ \\ \Delta H_f^0 of H_{2}O\left(l\right)=-286.0kJ \\ \Delta H_f^0 of C_2{H}_{5}OH\left(l\right)=-278.0kJ \\ \Delta H_f^0 of O_2\left(g\right)=0kJ \end{gathered}$$

Determination of standard enthalpy for the combustion of ethanol

The standard enthalpy for combustion of ethanol is,

$$\begin{gathered} \Delta H^o=\Delta H_f^o\left(product\right)-\Delta H_f^o\left(reactant\right) \\ \Delta H^o=\left[2\times \Delta H_f^o\left(C{O}_2\right)+3\times \Delta H_f^o\left(H_{2}O\right)\right]-\left[\Delta H_f^o\left(C_2{H}_{5}OH\right)+3\times \Delta H_f^o\left(O_2\right)\right] \\ \Delta H^o=\left[2\times \left(-393.5kJ\right)+3\times \left(-286.0kJ\right)\right]-\left[\left(-278.0kJ\right)+\left(3\times 0\right)\right] \\ \Delta H^o=-1367.0kJ/mol \end{gathered}$$

1 mol of ethanol = 46 g/mol

Heat released per 46 g of ethanol = -1367 kJ

Heat released per gram of ethanol is,

$$\begin{gathered} =\frac{-1367.0kJ}{46g} \\ =-29.71kJ/g \end{gathered}$$

Thus, the standard enthalpy for combustion of ethanol is -29.71 kJ/g.