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The standard heat of formation, [tex]$\Delta H_{ f }^{\circ}$[/tex], is defined as the enthalpy change for the formation of one mole of substance from its constituent elements in their standard states. Thus, elements in their standard states have [tex]$\Delta H_{ f }^{\circ}=0$[/tex]. Heat of formation values can be used to calculate the enthalpy change of any reaction.

Consider, for example, the reaction

[tex]\[ 2 NO (g) + O_2(g) \rightleftharpoons 2 NO_2(g) \][/tex]

with heat of formation values given by the following table:

[tex]\[
\begin{tabular}{|c|c|}
\hline
Substance & \begin{tabular}{c}
$\Delta H_{ f }^{\circ}$ \\
$(kJ / mol)$
\end{tabular} \\
\hline
$NO (g)$ & 90.2 \\
\hline
$O_2(g)$ & 0 \\
\hline
$NO_2(g)$ & 33.2 \\
\hline
\end{tabular}
\][/tex]

Then the standard heat of reaction for the overall reaction is

[tex]\[
\begin{aligned}
\Delta H_{ \text{rxn} }^{\circ} &= \Delta H_{ f }^{\circ}(\text {products}) - \Delta H_{ f }^{\circ} (\text{reactants}) \\
&= [2 \times 33.2 - (2 \times 90.2 + 0)] \\
&= -114 \, \text{kJ}
\end{aligned}
\][/tex]

Part B

The combustion of propane, [tex]C_3 H_8[/tex], occurs via the reaction

[tex]\[ C_3 H_8(g) + 5 O_2(g) \rightarrow 3 CO_2(g) + 4 H_2O(g) \][/tex]

with heat of formation values given by the following table:

[tex]\[
\begin{tabular}{|c|c|}
\hline
Substance & \begin{tabular}{c}
$\Delta H_f$ \\
$(kJ / mol)$
\end{tabular} \\
\hline
$C_3 H_8(g)$ & -104.7 \\
\hline
$CO_2(g)$ & -393.5 \\
\hline
$H_2O(g)$ & -241.8 \\
\hline
\end{tabular}
\][/tex]

Calculate the enthalpy for the combustion of 1 mole of propane.
Express your answer to four significant figures and include the appropriate units.

[tex]\[ \Delta H_{\text{rxn}}^{\circ} = \, \square \, \text{Value} \, \text{Units} \][/tex]



Answer :

GinnyAnswer
To calculate the enthalpy change ([tex]\(\Delta H_{rxn}^{\circ}\)[/tex]) for the combustion of propane ([tex]\(C_3H_8\)[/tex]), we need to apply the principle of using standard heats of formation ([tex]\(\Delta H_f^{\circ}\)[/tex]) of the reactants and products. Here is a step-by-step solution:

1. Write the balanced chemical equation for the combustion of propane:
[tex]\[ \mathrm{C_3H_8(g)} + 5 \mathrm{O_2(g)} \rightarrow 3 \mathrm{CO_2(g)} + 4 \mathrm{H_2O(g)} \][/tex]

2. List the given heat of formation values:
[tex]\[ \begin{array}{|c|c|} \hline \text{Substance} & \Delta H_f^{\circ} \, (\text{kJ/mol}) \\ \hline \mathrm{C_3H_8(g)} & -104.7 \\ \hline \mathrm{CO_2(g)} & -393.5 \\ \hline \mathrm{H_2O(g)} & -241.8 \\ \hline \mathrm{O_2(g)} & 0 \quad (\text{element in standard state}) \\ \hline \end{array} \][/tex]

3. Calculate the total enthalpy change for the products:
[tex]\[ \Delta H_{products}^{\circ} = (3 \times \Delta H_f^{\circ}(\mathrm{CO_2})) + (4 \times \Delta H_f^{\circ}(\mathrm{H_2O})) \][/tex]
Substituting the values:
[tex]\[ \Delta H_{products}^{\circ} = (3 \times -393.5) + (4 \times -241.8) \][/tex]
[tex]\[ \Delta H_{products}^{\circ} = -1180.5 + (-967.2) = -2147.7 \, \text{kJ} \][/tex]

4. Calculate the total enthalpy change for the reactants:
[tex]\[ \Delta H_{reactants}^{\circ} = (1 \times \Delta H_f^{\circ}(\mathrm{C_3H_8})) + (5 \times \Delta H_f^{\circ}(\mathrm{O_2})) \][/tex]
Substituting the values:
[tex]\[ \Delta H_{reactants}^{\circ} = (1 \times -104.7) + (5 \times 0) \][/tex]
[tex]\[ \Delta H_{reactants}^{\circ} = -104.7 \, \text{kJ} \][/tex]

5. Calculate the enthalpy change of the reaction ([tex]\(\Delta H_{rxn}^{\circ}\)[/tex]):
[tex]\[ \Delta H_{rxn}^{\circ} = \Delta H_{products}^{\circ} - \Delta H_{reactants}^{\circ} \][/tex]
Substituting the values:
[tex]\[ \Delta H_{rxn}^{\circ} = -2147.7 - (-104.7) \][/tex]
[tex]\[ \Delta H_{rxn}^{\circ} = -2147.7 + 104.7 = -2043.0 \, \text{kJ} \][/tex]

6. Express the final answer with appropriate units:
[tex]\[ \Delta H_{rxn}^{\circ} = -2043.0 \, \text{kJ} \][/tex]

Therefore, the enthalpy change for the combustion of 1 mole of propane is [tex]\(-2043.0 \, \text{kJ}\)[/tex].

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