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

(a) \({{\rm{H}}_{\rm{2}}}{\rm{(g) + B}}{{\rm{r}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2HBr(g)}}\)

(b) \({\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{(g) + }}{{\rm{I}}_{\rm{2}}}{\rm{(g)}} \to {\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{I(g) + HI(g)}}\)

(c) \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_4}{\rm{(g) + 3}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2C}}{{\rm{O}}_{\rm{2}}}{\rm{(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\)

Short Answer

Expert verified

(a) For the reaction \({{\rm{H}}_{\rm{2}}}{\rm{(g) + B}}{{\rm{r}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2HBr(g)}}\), the value for enthalpy change is \({\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = - 144\;kJ}}\).

(b) For the reaction\({\rm{C}}{{\rm{H}}_{\rm{4}}}{\rm{(g) + }}{{\rm{I}}_{\rm{2}}}{\rm{(g)}} \to {\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{I(g) + HI(g)}}\), the value for enthalpy change is\({\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = 30\;kJ}}\).

(c) For the reaction \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_4}{\rm{(g) + 3}}{{\rm{O}}_{\rm{2}}}{\rm{(g)}} \to {\rm{2C}}{{\rm{O}}_{\rm{2}}}{\rm{(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\), the value for enthalpy change is \({\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = - 1055\;kJ}}\).

Step by step solution

01

Concept Introduction

Bond energy, also known as mean bond enthalpy or average bond enthalpy in chemistry, is a measure of the bond strength in a chemical bond. The average value of the gas-phase bond-dissociation energy for all bonds of the same type within the same chemical species is defined by IUPAC as bond energy.

02

Enthalpy Change for first reaction

(a)

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

Bond Energy for\({\rm{Br - Br}}\)is:\({\rm{190 kJ/mol}}\)

Bond Energy for\({\rm{H - Br}}\)is:\({\rm{370 kJ/mol}}\)

Calculate the enthalpy change according to the reaction –

\(\begin{array}{l}{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = }}\sum {{{\rm{D}}_{{\rm{bonds broken }}}}} {\rm{ - }}\sum {{{\rm{D}}_{{\rm{bonds formed }}}}} \\{\rm{\Delta }}{{\rm{H}}^{\rm{^\circ }}}_{{\rm{298}}}{\rm{ = }}{{\rm{D}}_{{\rm{H - H}}}}{\rm{ + }}{{\rm{D}}_{{\rm{Br - Br}}}}{\rm{ - 2}}{{\rm{D}}_{{\rm{H - Br}}}}\\{\rm{\Delta }}{{\rm{H}}^{\rm{^\circ }}}_{{\rm{298}}}{\rm{ = 436 + 190 - 2(370) = - 144\;kJ}}\end{array}\)

Therefore, the enthalpy change is \({\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = - 144\;kJ}}\).

03

Enthalpy Change for second reaction

(b)

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

Bond Energy for\({\rm{I - I}}\)is:\({\rm{150 kJ/mol}}\)

Bond Energy for\({\rm{C - I}}\)is:\({\rm{240 kJ/mol}}\)

Bond Energy for\({\rm{H - I}}\)is:\({\rm{295 kJ/mol}}\)

Calculate the enthalpy change according to the reaction –

\(\begin{array}{l}{\rm{\Delta }}{{\rm{H}}^{\rm{^\circ }}}_{{\rm{298}}}{\rm{ = }}\sum {{D_{{\rm{bonds broken }}}}} {\rm{ - }}\sum {{D_{{\rm{bonds formed }}}}} \\{\rm{\Delta }}{{\rm{H}}^{\rm{^\circ }}}_{{\rm{298}}}{\rm{ = 4}}{{\rm{D}}_{{\rm{C - H}}}}{\rm{ + }}{{\rm{D}}_{{\rm{I - I}}}}{\rm{ - 3}}{{\rm{D}}_{{\rm{C - H}}}}{\rm{ - }}{{\rm{D}}_{{\rm{C - I}}}}{\rm{ - }}{{\rm{D}}_{{\rm{H - I}}}}\\{\rm{\Delta }}{{\rm{H}}^{\rm{^\circ }}}_{{\rm{298}}}{\rm{ = 4(415) + 150 - 3(415) - 240 - 295 = 30\;kJ}}\end{array}\)

Therefore, the enthalpy change is \({\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = 30\;kJ}}\).

04

Enthalpy Change for third reaction

(c)

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

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

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 –

\(\begin{array}{l}{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = }}\sum {{{\rm{D}}_{{\rm{bonds broken }}}}} {\rm{ - }}\sum {{{\rm{D}}_{{\rm{bonds formed }}}}} \\{\rm{\Delta }}{{\rm{H}}^{\rm{^\circ }}}_{298}{\rm{ = }}{{\rm{D}}_{{\rm{C = C}}}}{\rm{ + 4}}{{\rm{D}}_{{\rm{C - H}}}}{\rm{ + 3}}{{\rm{D}}_{{\rm{O = O}}}}{\rm{ - 4}}{{\rm{D}}_{{\rm{C = O}}}}{\rm{ - 4}}{{\rm{D}}_{{\rm{O - H}}}}\\{\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = 611 + 4(415) + 3(498) - 4(741) - 4(464) = - 1055\;kJ}}\end{array}\)

Therefore, the enthalpy change is \({\rm{\Delta H}}_{{\rm{298}}}^{\rm{^\circ }}{\rm{ = - 1055\;kJ}}\).

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