Question: Using the bond energies in Table \({\rm{7}}{\rm{.2}}\), determine the approximate enthalpy change for each of the following reactions:

(a) \({\rm{C}}{{\rm{l}}_{\rm{2}}}{\rm{(g) + 3}}{{\rm{F}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2Cl}}{{\rm{F}}_{\rm{3}}}{\rm{(g)}}\)

(b) \({{\rm{H}}_{\rm{2}}}{\rm{C = C}}{{\rm{H}}_{\rm{2}}}{\rm{(g) + }}{{\rm{H}}_{\rm{2}}}{\rm{(g)}} \to {{\rm{H}}_{\rm{3}}}{\rm{CC}}{{\rm{H}}_{\rm{3}}}{\rm{(g)}}\)

(c) \({\rm{2}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{\rm{(g) + 7}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{4C}}{{\rm{O}}_{\rm{2}}}{\rm{(g) + 6}}{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\)

Bond energy is a measure of the bond strength in a chemical bond. The average value of bond-dissociation energy for all bonds of the same type in gas phase within the same chemical species gives the value of bond energy.

(a)

Bond Energy for\({\rm{Cl - Cl}}\)is:\({\rm{243 kJ/mol}}\)

Bond Energy for\({\rm{F - F}}\)is:\({\rm{160 kJ/mol}}\)

Bond Energy for \({\rm{Cl - F}}\) is: \({\rm{255 kJ/mol}}\)

Calculate the enthalpy change according to the reaction –

\(\begin{array}{c}{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = }}\sum {{{\rm{D}}_{{\rm{bonds broken }}}}} {\rm{ - }}\sum {{{\rm{D}}_{{\rm{bonds formed }}}}} \\{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = 2}}{{\rm{D}}_{{\rm{Cl - Cl}}}}{\rm{ + 3}}{{\rm{D}}_{{\rm{F - F}}}}{\rm{ - 6}}{{\rm{D}}_{{\rm{Cl - F}}}}\\{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = }}2(243) + 3(160) - 6(255) = {\rm{ - 564\;kJ}}\end{array}\)

Therefore, the enthalpy change is \({\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = - 564\;kJ}}\).

(b)

Bond Energy for\({\rm{C = C}}\)is:\({\rm{611 kJ/mol}}\)

Bond Energy for\({\rm{C - H}}\)is:\({\rm{415 kJ/mol}}\)

Bond Energy for\({\rm{H - H}}\)is:\({\rm{436 kJ/mol}}\)

Bond Energy for\({\rm{C - C}}\)is:\({\rm{345 kJ/mol}}\)

Calculate the enthalpy change according to the reaction –

\(\begin{array}{l}{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = }}\sum {{{\rm{D}}_{{\rm{bonds broken }}}}} {\rm{ - }}\sum {{{\rm{D}}_{{\rm{bonds formed }}}}} \\{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = }}{{\rm{D}}_{{\rm{C = C}}}}{\rm{ + 4}}{{\rm{D}}_{{\rm{C - H}}}}{\rm{ + }}{{\rm{D}}_{{\rm{H - H}}}}{\rm{ - }}{{\rm{D}}_{{\rm{C - C}}}}{\rm{ - 6}}{{\rm{D}}_{{\rm{C - H}}}}\\{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = 611 + 4(415) + 436 - 345 - 6(415) = - 128\;kJ}}\end{array}\)

Therefore, the enthalpy change is \({\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = - 128\;kJ}}\).

(c)

Bond Energy for\({\rm{C - C}}\)is:\({\rm{345 kJ/mol}}\)

Bond Energy for\({\rm{C - H}}\)is:\({\rm{415 kJ/mol}}\)

Bond Energy for\({\rm{O = O}}\)is:\({\rm{498 kJ/mol}}\)

Bond Energy for\({\rm{C = O}}\)is:\({\rm{741 kJ/mol}}\)

Bond Energy for\({\rm{O - H}}\)is:\({\rm{464 kJ/mol}}\)

Calculate the enthalpy change according to the reaction –

\(\begin{array}{l}{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = }}\sum {{{\rm{D}}_{{\rm{bonds broken }}}}} {\rm{ - }}\sum {{{\rm{D}}_{{\rm{bonds formed }}}}} \\{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = 2}}{{\rm{D}}_{{\rm{C - C}}}}{\rm{ + 12}}{{\rm{D}}_{{\rm{C - H}}}}{\rm{ + 7}}{{\rm{D}}_{{\rm{O = O}}}}{\rm{ - 8}}{{\rm{D}}_{{\rm{C = O}}}}{\rm{ + 12}}{{\rm{D}}_{{\rm{O - H}}}}\\{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = 2(345) + 12(415) + 7(498) - 8(741) - 12(464) = - 2345\;kJ}}\end{array}\)

Therefore, the enthalpy change is \({\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = - 2345\;kJ}}\).