For a particular isomer of C8H18,C8H18, the combustion reaction produces 5093.7 kJ5093.7 kJ of heat per mole of C8H18(g)C8H18(g) consumed, under standard conditions.
C8H18(g)+252O2(g)?8CO2(g)+9H2O(g) ?H°rxn=-5093.7 kJ/mol
What is the standard enthalpy of formation of this isomer of C8H18(g)?
The correct answer and explanation is :
Let’s solve your question carefully.
You are given:
[
\text{C}8\text{H}{18}(g) + \frac{25}{2}\text{O}2(g) \rightarrow 8\text{CO}_2(g) + 9\text{H}_2\text{O}(g) \quad \Delta H^\circ{\text{rxn}} = -5093.7\ \text{kJ/mol}
]
You are asked to find the standard enthalpy of formation ((\Delta H^\circ_f)) of the isomer of (C_8H_{18}(g)).
Step 1: Recall the Enthalpy of Reaction Formula
The standard enthalpy change of a reaction is:
[
\Delta H^\circ_{\text{rxn}} = \sum \Delta H^\circ_f\ (\text{products}) – \sum \Delta H^\circ_f\ (\text{reactants})
]
For the combustion:
- Products: (8\ \text{CO}_2(g)) and (9\ \text{H}_2\text{O}(g))
- Reactants: (1\ \text{C}8\text{H}{18}(g)) and (\frac{25}{2}\ \text{O}_2(g))
Note: The standard enthalpy of formation for (O_2(g)) is zero.
Step 2: Known Values
From standard tables:
- (\Delta H^\circ_f\ \text{of}\ \text{CO}_2(g) = -393.5\ \text{kJ/mol})
- (\Delta H^\circ_f\ \text{of}\ \text{H}_2\text{O}(g) = -241.8\ \text{kJ/mol})
Step 3: Write the Equation
[
-5093.7\ \text{kJ} = [8 \times (-393.5) + 9 \times (-241.8)] – [\Delta H^\circ_f\ \text{of}\ \text{C}8\text{H}{18}]
]
First, calculate the products’ side:
[
8 \times (-393.5) = -3148.0\ \text{kJ}
]
[
9 \times (-241.8) = -2176.2\ \text{kJ}
]
Adding them:
[
-3148.0 + (-2176.2) = -5324.2\ \text{kJ}
]
Now plug into the equation:
[
-5093.7 = (-5324.2) – (\Delta H^\circ_f\ \text{of}\ \text{C}8\text{H}{18})
]
Step 4: Solve for (\Delta H^\circ_f)
Rearranging:
[
\Delta H^\circ_f\ \text{of}\ \text{C}8\text{H}{18} = -5324.2 + 5093.7
]
[
\Delta H^\circ_f\ \text{of}\ \text{C}8\text{H}{18} = -230.5\ \text{kJ/mol}
]
Final Answer:
[
\boxed{-230.5\ \text{kJ/mol}}
]
Explanation (~300 words)
In thermochemistry, the standard enthalpy of formation ((\Delta H^\circ_f)) of a compound is the heat change when one mole of the substance forms from its elements in their standard states. For hydrocarbons like octane ((C_8H_{18})), combustion is a convenient method to calculate it because the combustion enthalpy ((\Delta H^\circ_{\text{rxn}})) is typically measured experimentally.
The given reaction shows that when 1 mole of gaseous (C_8H_{18}) combusts, it releases (-5093.7\ \text{kJ}) of heat. In combustion, the products are typically (CO_2(g)) and (H_2O(g)), both of which have well-known standard enthalpies of formation.
The general formula we apply is:
[
\Delta H^\circ_{\text{rxn}} = \text{Sum of} (\Delta H^\circ_f\ \text{of products}) – \text{Sum of} (\Delta H^\circ_f\ \text{of reactants})
]
Oxygen ((O_2)) being an element in its standard state has (\Delta H^\circ_f = 0), simplifying the calculation. We multiply the known (\Delta H^\circ_f) values by their stoichiometric coefficients (8 for (CO_2), 9 for (H_2O)) and add them. This gives the total enthalpy change for the products.
By plugging in and rearranging the formula, we isolate the unknown (\Delta H^\circ_f) for octane. The negative value (-230.5\ \text{kJ/mol}) indicates that the formation of this isomer of octane from its elements is an exothermic process (it releases energy).
Understanding these relationships helps chemists predict reaction behavior and design fuels with desirable energy properties.