To calculate the reaction enthalpy ([tex]\(\Delta H^\circ_{\text{reaction}}\)[/tex]) for the given chemical equation under standard conditions:
[tex]\[ \text{C}_2\text{H}_2(g) + 2\text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g) \][/tex]
we will use the standard enthalpies of formation [tex]\(\Delta H^\circ_f\)[/tex] of the reactants and products.
First, list the standard enthalpies of formation ([tex]\(\Delta H^\circ_f\)[/tex]) from the ALEKS Data tab:
- [tex]\(\Delta H^\circ_f\)[/tex] for [tex]\(\text{C}_2\text{H}_2(g)\)[/tex]: 227 kJ/mol
- [tex]\(\Delta H^\circ_f\)[/tex] for [tex]\(\text{H}_2(g)\)[/tex]: 0 kJ/mol
- [tex]\(\Delta H^\circ_f\)[/tex] for [tex]\(\text{C}_2\text{H}_6(g)\)[/tex]: -84.7 kJ/mol
The reaction enthalpy [tex]\(\Delta H^\circ_{\text{reaction}}\)[/tex] can be calculated using the formula:
[tex]\[ \Delta H^\circ_{\text{reaction}} = \sum \Delta H^\circ_f(\text{products}) - \sum \Delta H^\circ_f(\text{reactants}) \][/tex]
Next, we'll calculate the sums of the enthalpies of formation for the reactants and the products:
1. Enthalpy of formation for the products:
- There is 1 mole of [tex]\(\text{C}_2\text{H}_6(g)\)[/tex]:
[tex]\[ \text{Sum of enthalpies of products} = 1 \times (-84.7) = -84.7 \text{ kJ/mol} \][/tex]
2. Enthalpy of formation for the reactants:
- There is 1 mole of [tex]\(\text{C}_2\text{H}_2(g)\)[/tex] and 2 moles of [tex]\(\text{H}_2(g)\)[/tex]:
[tex]\[ \text{Sum of enthalpies of reactants} = 1 \times 227 + 2 \times 0 = 227 \text{ kJ/mol} \][/tex]
3. Calculate the reaction enthalpy:
[tex]\[ \Delta H^\circ_{\text{reaction}} = \text{Sum of enthalpies of products} - \text{Sum of enthalpies of reactants} \][/tex]
[tex]\[ \Delta H^\circ_{\text{reaction}} = -84.7 \text{ kJ/mol} - 227 \text{ kJ/mol} = -311.7 \text{ kJ/mol} \][/tex]
Finally, round the answer to the nearest kJ:
[tex]\[ \Delta H^\circ_{\text{reaction}} \approx -312 \text{ kJ/mol} \][/tex]
Therefore, the reaction enthalpy for the given reaction is [tex]\(-312\)[/tex] kJ/mol.
