4.3: Question 4.E.65 PASS - determining enthalpy change from bond energies

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Exercise \(\PageIndex{4.E.65}\)

Using the bond energies in Table 4.6.1, determine the approximate enthalpy change for each of the following reactions:

1. \(\ce{H2}(g)+\ce{Br2}(g)⟶\ce{2HBr}(g)\)

2. \(\ce{CH4}(g)+\ce{I2}(g)⟶\ce{CH3I}(g)+\ce{HI}(g)\)

3. \(\ce{C2H4}(g)+\ce{3O2}(g)⟶\ce{2CO2}(g)+\ce{2H2O}(g)\)

Answer

1. −114 kJ

2. 30. kJ

3. −1055 kJ

See LibreText 4.6: Strengths of Ionic and Covalent Bonds (section 4.6.1 Bond Strength: Covalent Bonds)

Strategy Map

Step	Hint
1. Identify which molecules have bonds that are being broken and which have bonds that are being formed.	Reactant bonds are broken, product bonds are formed in a chemical reaction.
2. Count up the bonds broken and formed. If you do not know what the bonds are create Lewis structures for each of the molecules.	Recall how to draw Lewis Structures (see LibreText section 4.4)
3. Use the provided table to look up the corresponding bond energies.	Table 4.6.1
4. Plug the values into the relationship that calculates enthalpy change from bond energies and solve.	To break bonds an energy input is required, when bonds formed energy is released. When energy is released it is negative. Recall the equation that demonstrates this relationship.

Solution

Reaction 1:

\(\ce{H2}(g)+\ce{Br2}(g)⟶\ce{2HBr}(g)\)

\(\Delta H=\sum \mathrm{D}_{\text {bonds broken }}-\sum \mathrm{D}_{\text {bonds formed }}\)

\(\Delta H=\sum \mathrm{D}_{\text {reactant bonds}}-\sum \mathrm{D}_{\text {product bonds}}\)

\(\begin{gathered}
\Delta H=(1 mol(H-H)+1 mol(B r-B r))-(2 mol(H-B r)) \\
\Delta H=(1 mol(436 kJ/mol)+1 mol(190. kJ/mol))-(2mol(370. kJ/mol)) \\
\Delta H=(626 kJ)-(740. kJ) = -114 kJ
\end{gathered}\)

Answer -114 kJ

Guided Solution Reaction 1

Download Guided Solution as a pdf

Guided Solution	Hint
This is a calculation type problem in which you must identify what bonds are being broken and formed during a chemical reaction. This type of problem requires you to use a table of values to calculate the overall energy change in the reaction using bond energies.	See LibreText 4.6: Strengths of Ionic and Covalent Bonds (section 4.6.1 Bond Strength: Covalent Bonds)
Using the bond energies in Table 4.6.1, determine the approximate enthalpy change for each of the following reactions: 1. \(\ce{H2}(g)+\ce{Br2}(g)⟶\ce{2HBr}(g)\)
It can be difficult to know how many bonds are in a molecule. The easiest way to find what bonds there are is to create a Lewis Structure of the molecule.	Recall how to draw Lewis Structures (see LibreText section 4.4) H₂ Lewis Structure Br₂Lewis Structure HBr Lewis Structure
When using the bond energies table, scan the table for your desired bond, i.e. (H-H), the number beside it will be the one used in the calculation. Make sure you don’t confuse single, double, and triple bonds!	The bond energy for H-H is 436 kJ/mol.
Pay attention to stoichiometry! Multiply the stoichiometric coefficient by your bond energy.	Students often make mistakes with this step. Ensure you are multiplying your bond energy value by the stoichiometric coefficient as well as the number of times that bond shows up in the molecule.
Recall that it requires energy to break a bond (reactants) and energy is released when bonds are formed (products of a reaction). Remembering this should help you set up your equation to solve for \(\Delta\)H!	\(\Delta H=\sum \mathrm{D}_{\text {bonds broken }}-\sum \mathrm{D}_{\text {bonds formed }}\) \(\Delta H=\sum \mathrm{D}_{\text {reactant bonds}}-\sum \mathrm{D}_{\text {product bonds}}\) The energy of bonds formed is subtracted as they are formed in the reaction, and bond energy values by definition are the energy required to break the bond. By subtracting the sum of the product bonds formed we are changing the sign of the energy to represent the opposite process (forming the bond).
Complete Solution: The following equation will be used: \(\Delta H=\sum \mathrm{D}_{\text {bonds broken }}-\sum \mathrm{D}_{\text {bonds formed }}\) For a reaction to occur, bonds need to break, and new bonds need to form. When bonds break energy is absorbed. When bonds are formed, energy is released. We use this to calculate the sum of the energy change over the reaction. How do we know which bonds are breaking and which are forming? Identify and track your bonds: \(\ce{H2}(g)+\ce{Br2}(g)⟶\ce{2HBr}(g)\) The bond between the two hydrogen atoms and the two bromine atoms are broken. Two product molecules are formed each with a hydrogen bonded to a bromine. \(\Delta H=(1 mol(H-H)+1 mol(B r-B r))-(2mol(H-B r))\) Look up the bond energies, enter them in equation: \(\Delta H=(1 mol(436 kJ/mol)+1 mol(190. kJ/mol))-(2mol(370. kJ/mol)) \\\) Solve inside the brackets, remember stoichiometry: \(\Delta H=(626 kJ)-(740. kJ) \) The energy released from the bonds formed are subtracted from the energy absorbed by the bonds broken. \(\Delta H=(626 kJ)-(740. kJ) = -114 kJ\) The calculated value is negative, meaning the reaction was exothermic. Answer -114 kJ	In a reaction, the bonds to the left of the arrow will break and the bonds to the right of the arrow will form. ___(reactant bonds break)___ ⟶ ___(product bonds form)___ The bonds in these molecules are all single bonds. H-H 436 kJ/mol Br-Br 190. kJ/mol H-Br 370. kJ/mol Stoichiometric coefficients are exact numbers with infinite significant figures, so for H₂ is 1.000000..., so it is the bond energy values that determine significant figures for answer. You could also leave units of kJ/mol for the answer, since this represents kJ/mol of the reaction as written.
Check your work! Comparing the bond energy values we see that the sum of the product bond (2 H-Br) have a greater magnitude than the sum of reactant bonds (H-H and Br-Br) so we would expect this reaction to have an exothermic \(\Delta\)H. Why does this answer make chemical sense? When any reaction occurs, energy will be both absorbed and released, the overall sum however can be endothermic (requires more energy than it releases) or exothermic (it releases more energy than it requires). We can numerically observe this using the enthalpy change from bond energies equation.	Make sure you add up all bonds and consider the reaction stoichiometry.