Question

Using the data provided, calculate the multiple bond energy ( ${\text{kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$ ) of ${\text{C}} \equiv {\text{C}}$ bond in ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}$ . That energy is (take the bond energy of a ${\text{C - H}}$ bond as 350 ${\text{kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$ ):
${\text{2C }}\left( {\text{s}} \right){\text{ + }}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right) \to {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\left( {\text{g}} \right)$ ${{\Delta H = 225 kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$
${\text{2C }}\left( {\text{s}} \right) \to 2{\text{C}}\left( {\text{g}} \right)$ ${{\Delta H = 1410 kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$
${{\text{H}}_{\text{2}}} \to {\text{2 H}}\left( {\text{g}} \right)$ ; ${{\Delta H = 330 kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$
A.1165
B.837
C.865
D.815

Answer 1

The bond energy or the bond enthalpy in chemistry is defined as the measure of the bond strength of a chemical bond. The IUPAC has defined the bond energy to be equal to the gas phase bond dissociation energy. We shall calculate the enthalpy of reactants from the first reaction and then calculate the carbon-carbon triple bond energy.
Formula Used:
${{\Delta H = }}{{\text{H}}_{\text{r}}}{\text{ - }}{{\text{H}}_{\text{p}}}$
Where ${{\text{H}}_{\text{p}}}$ is the enthalpy of the products formed and ${{\text{H}}_{\text{r}}}$ is the enthalpy of the reactants.

Complete step by step answer:
The bond dissociation energy also referred to as the bond disruption energy, bond energy is the standard enthalpy change for the breaking of the bonds in one mole of molecules of a substance.
Since the enthalpy change of a reaction is equal to the difference between the enthalpy of the products and the enthalpy of the reactants so, mathematically it can be represented as,
${{\Delta H = }}{{\text{H}}_{\text{r}}}{\text{ - }}{{\text{H}}_{\text{p}}}$
Where ${{\text{H}}_{\text{p}}}$ is the enthalpy of the products formed which is equal to ${{\Delta }}{{\text{H}}_{{\text{C}} \equiv {\text{C}}}} + \left( {350 \times 2} \right)$ ${\text{kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$ , the bond energy for the formation of ethyne or acetylene.
And ${{\text{H}}_{\text{r}}}$ is the enthalpy of the reactants = ${\text{ }}\left( {{\text{1410 + 330}}} \right){\text{ kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$
Putting the values of the components in the above equation we get,
${{\Delta H = }}\left( {{\text{1410 + 330}}} \right){{ - \Delta }}{{\text{H}}_{{\text{C}} \equiv {\text{C}}}} + \left( {350 \times 2} \right)$
$\Rightarrow {\text{225 = }}\left( {{\text{1410 + 330}}} \right){{ - \Delta }}{{\text{H}}_{{\text{C}} \equiv {\text{C}}}} + \left( {350 \times 2} \right)$
Solving this:
$\Rightarrow {{\Delta }}{{\text{H}}_{{\text{C}} \equiv {\text{C}}}}{\text{ = 815}}$ ${\text{kJ mo}}{{\text{l}}^{{\text{ - 1}}}}$ .

Hence, the correct answer is option D.
Note:
The electronegativity of the atoms that hold together a covalent bond affect the bond dissociation energy. The size of the atoms involved in the bond as well as their position in the periodic table also affects the bond dissociation energies. As the electronegativity of fluorine is more than that of chlorine its bond dissociation energy should have been more than chlorine but as its size is smaller than chlorine, so due to electronic repulsion, the bond energy decreases.