Question

The standard enthalpy of formation (${ \Delta }_{ f }{ H }_{ 298 }^{ o }$) for methane, ${ CH }_{ 4 }$ is $-74.9\quad KJ{ mol }^{ -1 }$. In order to calculate the average energy given out in the formation of a C-H bond from this, it is necessary to know which one of the following?
(a) The dissociation energy of the hydrogen molecule ${ H }_{ 2 }$
(b) The first four ionization energy of carbon
(c) The dissociation energy of ${ H }_{ 2 }$ and the enthalpy of sublimation of carbon (graphite).
(d) The first four ionization energies of carbon and electron affinity of hydrogen

Answer 1

The enthalpy of formation of a substance is equal to the difference between the sum of the bond enthalpies of the reactants multiplied with their respective stoichiometric coefficient and the sum of the bond enthalpies of the products multiplied by their respective stoichiometric coefficients

Complete step by step solution:
In order to solve this question, we first need to understand the meaning of standard enthalpy of formation. The standard enthalpy of formation of a substance refers to the enthalpy change accompanying the formation of one mole of the substance in the standard state from its elements that are also taken in their standard state (at a temperature of 298 K and 1 bar pressure). Now, let us understand the meaning of bond energy. It is the amount of energy that is released when one mole of bonds are formed from the isolated atoms in the gaseous state or the amount of energy required to dissociate one mole of bonds present between the atoms in the gaseous molecules. For diatomic molecules, the bond energy is equal to their enthalpies of atomization. But for polyatomic molecules, the bond energy for a particular bond is not the same when it is present in different types of molecules or even in the same molecule. Hence in a molecule of methane, the bond energy for the first C-H bond is 427 KJ/mol, for the second C-H bond it is 439 KJ/mol and so on. Their sum will give us the enthalpy of atomisation of the methane molecule and when that value is divided by 4 (since methane has four C-H bonds) we will get the average bond energy.
Given below is the reaction for the formation of methane gas:
$C(s)+2{ H }_{ 2 }(g)\rightarrow { CH }_{ 4 }(g)$ (Here all the reactants and the products are taken in standard state and the carbon is actually graphite).
Now the standard enthalpy of formation of methane is equal to the difference between the sum of the bond enthalpies of the reactants multiplied with their respective stoichiometric coefficient and the sum of the bond enthalpies of the products multiplied by their respective stoichiometric coefficients i.e.
$\begin{matrix} \Delta { H }_{ f } \\\ Enthalpy\quad of \\\ formation\quad of\quad { CH }_{ 4 } \end{matrix}=\begin{matrix} (2\Delta { { H }^{ o } }_{ B(H-H) }+\Delta _{ sub }{ H }^{ o }(carbon)) \\\ sum\quad of\quad bond\quad enthalpy\quad and \\\ enthalpy\quad of\quad sublimation \end{matrix}-\begin{matrix} 4\Delta { { H }^{ o } }_{ B(C-H) } \\\ Bond\quad enthalpy \\\ of\quad C-H\quad bond \end{matrix}$
In the formula the standard bond enthalpy for carbon is replaced by its sublimation enthalpy since the value of the enthalpy of atomisation for carbon is equal to its sublimation enthalpy.

Hence the correct answer is (c) The dissociation energy of ${ H }_{ 2 }$ and the enthalpy of sublimation of carbon (graphite).

Note: The enthalpy of formation of a substance involves the formation of 1 mole of the substance from its elements under given conditions of temperature and pressure. Bond dissociation energy and bond dissociation enthalpy are actually different quantities. The bond dissociation energy values are measured at 0K using spectroscopic methods while the bond dissociation enthalpy values are calculated using heat capacities the pressure-volume work. But their values are close and hence are used interchangeably.