Question

In the complete combustion of butanol (${C_4}{H_9}OH$):
A. $\Delta H < \Delta E$
B. $\Delta H = \Delta E$
C. $\Delta H > \Delta E$
D. $\Delta H,\Delta E$ relation cannot be predicted

Answer 1

Hint: The complete combustion of butanol will give water and carbon dioxide as it is a hydrocarbon. In the options $\Delta H$ represents change in enthalpy and $\Delta E$ represents change in internal energy. The relation between change in enthalpy and internal energy change is given by the relation:
$\Delta H = \Delta E + \Delta {n_g}RT$
$\Delta H = $ Change in enthalpy
$\Delta E = $ Change in internal energy
$\Delta {n_g} = $ Number of gaseous molecules of products - number of gaseous molecules of reactants
$R = $ Universal gas constant
$T = $ Temperature

Complete step by step answer:
Combustion of butanol gives the following reaction:
${C_4}{H_9}OH(l) + 6{O_2}(g) \to 4C{O_2}(g) + 5{H_2}O(l)$
Combustion is a chemical process in which the substance reacts rapidly with oxygen and gives off heat. (One molecule of butanol reacts with six molecules of oxygen gas to form four molecules of carbon dioxide gas and five molecules of water)
Now we will calculate the number of gaseous molecules of reactants and products.
Number of gaseous molecules of reactants $ = 6$ (6 molecules of oxygen gas)
Number of gaseous molecules of products $ = 4$ (4 molecules of carbon dioxide gas)
$\Delta {n_g}$ (Number of gaseous molecules of products $ - $ number of gaseous molecules of reactants) $ = 4 - 6 = - 2$
Now putting in the above relation between change in enthalpy and internal energy $\Delta H = \Delta E + \Delta {n_g}RT$
$ \Rightarrow \Delta H = \Delta E - 2RT$ (Since number of gaseous molecules of products $ - $ number of gaseous molecules of reactants is $ - 2$)
$ \Rightarrow \Delta H + 2RT = \Delta E$
$ \Rightarrow \Delta E > \Delta H$

So, the correct answer is Option A.

Note: $\Delta {n_g}$ is always calculated for gaseous reactants and products only.
If $\Delta {n_g}$ is positive then $\Delta H > \Delta E$ (change in enthalpy is greater than change in internal energy).
If $\Delta {n_g}$ is negative then $\Delta H < \Delta E$ (change in enthalpy is less than change in internal energy).
If $\Delta {n_g} = 0$ then $\Delta H = \Delta E$ (change in enthalpy is equal to change in internal energy).