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Bromination of isobutane is a two-step reaction, as shown below. Using the table of bond dissociation energies, calculate the enthalpy of each step and the enthalpy of the overall reaction.
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Hess’s law says that the total change in enthalpy of a particular series of reactions is exactly similar to that change in the whole reaction. There are a variety of methods of applying Hess’s law. The first method is to employ an array of reactions that are added to the total equation of reactions. In this scenario, the variation in the enthalpy of the reactions is considered. Another way to apply Hess’s law is to apply the reactions that form for each product and reactant. If this is the case, then the temperatures of formation for each chemical are used. The final method is an approximation to Hess’s law , and will only provide some approximate figures. The final method employs an enzymatic bond break. One of the parameters used in this method is the energy of dissociation in bonds. This is the amount of energy required for breaking an average bond inside an exclusive molecule of gaz phase. The exact values for dissociation energy of bonds vary with each compound, however the average is able to be utilized.
This problem can be solved using the enthalpy-of-reaction concept.
The enthalpy of any reaction is the amount of energy required to cause the reaction. This refers to the change in enthalpy when the reactants react to form products. Bond dissociation energy is the energy needed to break a bond. It also refers to the energy that is released when forming a new bond. You may notice that both bond dissociation and bond formation occur in chemical reactions.
You can calculate the enthalpy for a reaction by using bond dissociation energie as follows.
[katex]\Delta H = \sum \Delta H\left( {{\rm{reactants}}} \right) – \sum \Delta H\left( {{\rm{products}}} \right)[/katex]
Here, [katex]\Delta H[/katex]
Part 1
You can read the following description of the reaction to acetylene combustion:
[katex]{\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{C}} – {\rm{H}} + {\rm{Br}} \to {\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{{\rm{C}}^ \bullet }{\rm{ + H}} – {\rm{Br}}[/katex]
This is how you can write the enthalpy of this reaction.
[katex]\Delta {H_1} = \Delta H\left( {{{\left( {{\rm{C}}{{\rm{H}}_3}} \right)}_3}{\rm{C}} – {\rm{H}}} \right) – \Delta H\left( {{\rm{H}} – {\rm{Br}}} \right)[/katex]
Substitute “equa_tag_4”
[katex]\begin{array}{c}\\\Delta {H_1} = \left( {400 – 366} \right){\rm{ kJ mo}}{{\rm{l}}^{ – 1}}\\\\ = + 34{\rm{ kJ mo}}{{\rm{l}}^{ – 1}}\\\end{array}[/katex]
Part 2
This is the reaction that causes acetylene to be combusted:
[katex]{\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{{\rm{C}}^ \bullet } + {\rm{Br}} – {\rm{Br}} \to {\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{C}} – {\rm{Br}} + {\rm{Br}}[/katex]
This is how you can write the enthalpy of this reaction:
[katex]\Delta {H_2} = \Delta H\left( {{\rm{Br}} – {\rm{Br}}} \right) – \Delta H\left( {{{\left( {{\rm{C}}{{\rm{H}}_3}} \right)}_3}{\rm{C}} – {\rm{Br}}} \right)[/katex]
Substitute “equa_tag_8”
[katex]\begin{array}{c}\\\Delta {H_2} = \left( {193 – 292} \right){\rm{ kJ mo}}{{\rm{l}}^{ – 1}}\\\\ = – 99{\rm{ kJ mo}}{{\rm{l}}^{ – 1}}\\\end{array}[/katex]
Part 3
This is the reaction that causes acetylene to be combusted:
[katex]{\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{C}} – {\rm{H}} + {\rm{Br}} – {\rm{Br}} \to {\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{C}} – {\rm{Br}} + {\rm{H}} – {\rm{Br}}[/katex]
Add the enthalpy from step 1 and 2 to calculate the enthalpy for overall reaction.
[katex]\Delta H = \Delta {H_1} + \Delta {H_2}[/katex]
Substitute “equa_tag_12”
[katex]\begin{array}{c}\\\Delta H = \left( {34 – 99} \right){\rm{ kJ mo}}{{\rm{l}}^{ – 1}}\\\\ = – 65{\rm{ kJ mo}}{{\rm{l}}^{ – 1}}\\\end{array}[/katex]Ans Part 1
The reaction’s enthalpy is [katex] + 34{\rm{ kJ mo}}{{\rm{l}}^{ – 1}}[/katex].