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

Consider, for example, the reaction
[tex]\[ 2 NO(g) + O_2(g) \rightarrow 2 NO_2(g) \][/tex]
with heat of formation values given by the following table:

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

Then the standard heat of reaction for the overall reaction is
[tex]\[
\begin{aligned}
\Delta H_{rxn} &= \Delta H_f \text{ (products)} - \Delta H_f \text{ (reactants)} \\
&= 2(33.2) - [2(90.2) + 0] \\
&= -113.6 \text{ kJ}
\end{aligned}
\][/tex]

Part A

For which of the following reactions is [tex]$\Delta H_{rxn}$[/tex] equal to [tex]$\Delta H_f$[/tex] of the product(s)? You do not need to look up any values to answer this question. Check all that apply.

A. [tex]$Na(s) + \frac{1}{2} F_2(g) \rightarrow NaF(s)$[/tex]

B. [tex]$CaCO_3(s) \rightarrow CaO(s) + CO_2(g)$[/tex]

C. [tex]$CO(g) + \frac{1}{2} O_2(g) \rightarrow CO_2(g)$[/tex]

D. [tex]$2 Na(s) + F_2(g) \rightarrow 2 NaF(s)$[/tex]

E. [tex]$Na(s) + \frac{1}{2} F_2(g) \rightarrow NaF(s)$[/tex]

F. [tex]$C(s,\ \text{graphite}) + O_2(g) \rightarrow CO_2(g)$[/tex]

Part B

The combustion of propene, [tex]$C_3H_6$[/tex], occurs via the reaction
[tex]\[ C_3H_6(g) + 5 O_2(g) \rightarrow 3 CO_2(g) + 4 H_2O(l) \][/tex]

With heat of formation values given by the following table:

[tex]\[
\begin{tabular}{|c|c|}
\hline
Substance & $\Delta H_f$ (kJ/mol) \\
\hline
C_3H_6(g) & -20.4 \\
\hline
O_2(g) & 0 \\
\hline
CO_2(g) & -393.5 \\
\hline
H_2O(l) & -285.8 \\
\hline
\end{tabular}
\][/tex]

Calculate the enthalpy change for the combustion of propene.

Sagot :

GinnyAnsweridn
Certainly! Let's work through this problem together.

### Part A:

We need to determine for which of the given reactions the enthalpy change of the reaction ([tex]\(\Delta H_{\text{rxn}}\)[/tex]) is equal to the heat of formation of the product(s) ([tex]\(\Delta H_f\)[/tex]).

For a reaction of the formation type, it must form one mole of product from its elements in their standard states.

Here are the given reactions:

1. [tex]\( \text{Na (s)} + \frac{1}{2} \text{F}_2 (\text{l}) \rightarrow \text{NaF (s)} \)[/tex]
2. [tex]\( \text{CaCO}_3 (\text{g}) \rightarrow \text{CaO} + \text{CO}_2 (\text{g}) \)[/tex]
3. [tex]\( \text{CO (g)} + \frac{1}{2} \text{O}_2 (\text{g}) \rightarrow \text{CO}_2 (\text{g}) \)[/tex]
4. [tex]\( 2 \text{Na (s)} + \text{F}_2 (\text{s}) + 2 \text{NaF (s)} \)[/tex]
5. [tex]\( \text{Na (s)} + \frac{1}{2} \text{F}_2 (\text{g}) \rightarrow \text{NaF (s)} \)[/tex]
6. [tex]\( \text{C (s, graphite)} + \text{O}_2 (\text{g}) \rightarrow \text{CO}_2 (\text{s}) \)[/tex]

Let's analyze each reaction:

1. Reaction 1: [tex]\( \text{Na (s)} + \frac{1}{2} \text{F}_2 (\text{l}) \rightarrow \text{NaF (s)} \)[/tex]
- Sodium and fluorine are in their standard states, and one mole of NaF is being formed. This is a formation reaction.

2. Reaction 2: [tex]\( \text{CaCO}_3 (\text{g}) \rightarrow \text{CaO} + \text{CO}_2 (\text{g}) \)[/tex]
- This is a decomposition reaction, not a formation reaction.

3. Reaction 3: [tex]\( \text{CO (g)} + \frac{1}{2} \text{O}_2 (\text{g}) \rightarrow \text{CO}_2 (\text{g}) \)[/tex]
- Carbon monoxide and oxygen are in their standard states, and one mole of CO[tex]\(_2\)[/tex] is being formed. This is a formation reaction.

4. Reaction 4: [tex]\( 2 \text{Na (s)} + \text{F}_2 (\text{s}) + 2 \text{NaF (s)} \)[/tex]
- This reaction does not make any sense as written and does not form NaF from elements in their standard states.

5. Reaction 5: [tex]\( \text{Na (s)} + \frac{1}{2} \text{F}_2 (\text{g}) \rightarrow \text{NaF (s)} \)[/tex]
- Sodium and fluorine are in their standard states, and one mole of NaF is being formed. This is a formation reaction.

6. Reaction 6: [tex]\( \text{C (s, graphite)} + \text{O}_2 (\text{g}) \rightarrow \text{CO}_2 (\text{s}) \)[/tex]
- Carbon (graphite) and oxygen are in their standard states, and one mole of CO[tex]\(_2\)[/tex] is being formed. This is a formation reaction.

Therefore, the reactions in which [tex]\(\Delta H_{\text{rxn}}\)[/tex] is equal to [tex]\(\Delta H_f\)[/tex] of the product(s) are reactions 1, 3, 5, and 6.

### Part B:

We are asked to calculate the enthalpy change for the combustion of propene ([tex]\(\text{C}_3\text{H}_6\)[/tex]) with the given reaction:

[tex]\[ \text{C}_3\text{H}_6 (\text{g}) + 5 \text{O}_2 (\text{g}) \rightarrow 3 \text{CO}_2 (\text{g}) + 4 \text{H}_2\text{O} (\text{g}) \][/tex]

Given the heat of formation values:
- [tex]\( \Delta H_f (\text{CO}_2 (\text{g})) = -393.5 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f (\text{H}_2\text{O} (\text{g})) = -241.8 \text{ kJ/mol} \)[/tex]
- [tex]\( \Delta H_f (\text{C}_3\text{H}_6) = \text{Not provided} \)[/tex]
- [tex]\( \Delta H_f (\text{O}_2 (\text{g})) = 0 \text{ kJ/mol} \)[/tex] (standard state)

The enthalpy change of reaction, [tex]\(\Delta H_{\text{rxn}}\)[/tex], is calculated as follows:

[tex]\[ \Delta H_{\text{rxn}} = \left[ 3 (\Delta H_f (\text{CO}_2 (\text{g}))) + 4 (\Delta H_f (\text{H}_2\text{O} (\text{g}))) \right] - \left[ 1 (\Delta H_f (\text{C}_3\text{H}_6)) + 5 (\Delta H_f (\text{O}_2 (\text{g}))) \right] \][/tex]

Substituting the given values:

[tex]\[ \Delta H_{\text{rxn}} = \left[ 3 (-393.5) + 4 (-241.8) \right] - \left[ 1 (\Delta H_f (\text{C}_3\text{H}_6)) + 5 (0) \right] \][/tex]

Since [tex]\(\Delta H_f (\text{C}_3\text{H}_6)\)[/tex] is not provided, the enthalpy change of the reaction, [tex]\(\Delta H_{\text{rxn}}\)[/tex], cannot be calculated exactly without this value.

So, we will leave the enthalpy change as undefined until the heat of formation for propene ([tex]\(\text{C}_3\text{H}_6\)[/tex]) is provided.

Therefore,

### Final Results:

- Part A: Reactions [tex]\( 1, 3, 5, \text{ and } 6 \)[/tex] are the ones for which [tex]\(\Delta H_{\text{rxn}}\)[/tex] is equal to [tex]\(\Delta H_f\)[/tex] of the product(s).
- Part B: The enthalpy change for the combustion of propene cannot be calculated without the heat of formation value for [tex]\(\text{C}_3\text{H}_6\)[/tex].
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