. Advertisement .
..3..
. Advertisement .
..4..
......... ADVERTISEMENT ......... ..8..
Use the molar bond enthalpy data in the table to estimate the value of ΔH0rmx for the equation.
Chemical reactions are the process of making and breaking of chemical bonds. Breaking chemical bonds requires the absorption of energy. The the formation of chemical bonds results to the release and discharge of energy. The enthalpy changes is calculated by calculating the bond energy using the following relationship: ΔHf=ΣEBond,products−ΣEBond,reactants Calculating the bond energies total of the bonds within the products , and subtracting the equivalent amount for the reactants, you can determine whether there is an net absorption or emission of energy, also known as the change in enthalpy.♦ Relevant knowledge
The general equation for heat of reaction can solve this problem.
Bond energy can be used to calculate the heat of a react or the change in enthalpy. After balancing the equation, heat energy can be estimated.
Bond energy can be defined as the energy required to break 1 mole of a gaseous substance to make its constituents.
The difference between the summation bond energies of bonds broken and the summation bond energies from the bonds formed is called the change in enthalpy. It can be read as follows.
[katex]{\rm{\Delta }}{H^o} = \sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}} – \sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}} [/katex]
This is the reaction:
[katex]{\rm{CC}}{{\rm{l}}_{\rm{4}}}\left( g \right) + 2{{\rm{F}}_2}\left( g \right) \to {\rm{C}}{{\rm{F}}_4}\left( g \right) + 2{\rm{C}}{{\rm{l}}_{\rm{2}}}\left( g \right)[/katex]
Calculate the bond energy of reactants and the bonds that have been broken first.
[katex]\sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}} = 4{\rm{\Delta }}H\left( {{\rm{C}} – {\rm{Cl}}} \right) + 2{\rm{\Delta }}H\left( {{\rm{F}} – {\rm{F}}} \right)[/katex]
Substitute [katex]{\rm{331 kJ mo}}{{\rm{l}}^{ – 1}}[/katex]
[katex]\begin{array}{c}\\\sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}} = 4\left( {{\rm{331 kJ mo}}{{\rm{l}}^{ – 1}}} \right) + 2\left( {{\rm{155 kJ mo}}{{\rm{l}}^{ – 1}}} \right)\\\\ = 1634{\rm{ kJ mo}}{{\rm{l}}^{ – 1}}\\\end{array}[/katex]
Calculate the bond energy for product bonds you have created.
[katex]\sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}} = {\rm{4\Delta }}H\left( {{\rm{C}} – {\rm{F}}} \right) + 2{\rm{\Delta }}H\left( {{\rm{Cl}} – {\rm{Cl}}} \right)[/katex]
Substitute [katex]{\rm{439 kJ mo}}{{\rm{l}}^{ – 1}}[/katex]
[katex]\begin{array}{c}\\\sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}} = 4\left( {{\rm{439 kJ mo}}{{\rm{l}}^{ – 1}}} \right) + 2\left( {{\rm{243 kJ mo}}{{\rm{l}}^{ – 1}}} \right)\\\\ = 2242{\rm{ kJ mo}}{{\rm{l}}^{ – 1}}\\\end{array}[/katex]
Substitute 1634 kJ now for [katex]\sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}} [/katex]
[katex]\begin{array}{c}\\{\rm{\Delta }}{H^o} = \sum {{\rm{\Delta }}{H_{\left( {{\rm{reactant bonds broken}}} \right)}}} – \sum {{\rm{\Delta }}{H_{\left( {{\rm{product bonds formed}}} \right)}}} \\\\ = \left( {1634 – 2242} \right){\rm{ kJ mo}}{{\rm{l}}^{ – 1}}\\\\ = – 608{\rm{ kJ mo}}{{\rm{l}}^{ – 1}}\\\end{array}[/katex]Ans:
[katex]{\rm{CC}}{{\rm{l}}_{\rm{4}}}\left( g \right) + 2{{\rm{F}}_2}\left( g \right) \to {\rm{C}}{{\rm{F}}_4}\left( g \right) + 2{\rm{C}}{{\rm{l}}_{\rm{2}}}\left( g \right)[/katex]0 is the heat of the reaction