The combustion of octane is shown in the unbalanced reaction below _____C8H18(l) + _____O2(g) à _____CO2(g) + _____H2O(g) What is the enthalpy of reaction for the combustion of 1 mol of octane?

To determine the enthalpy of reaction for the combustion of 1 mole of octane, we need to follow these steps:

### 1. **Balance the Chemical Equation:**

The unbalanced equation is:
\[
\text{C}_8\text{H}_{18}(l) + \text{O}_2(g) \rightarrow \text{CO}_2(g) + \text{H}_2\text{O}(g)
\]

Balance the equation:

- **Carbon (C):** Octane has 8 carbon atoms, so you need 8 CO₂ molecules.
- **Hydrogen (H):** Octane has 18 hydrogen atoms, so you need 9 H₂O molecules.
- **Oxygen (O):** The products have \(8 \times 2 = 16\) oxygen atoms from CO₂ and \(9 \times 1 = 9\) oxygen atoms from H₂O, totaling 25 oxygen atoms. Since O₂ is diatomic, you need \(\frac{25}{2}\) molecules of O₂, which is not an integer. To avoid fractions, multiply the entire equation by 2.

The balanced equation becomes:
\[
2\text{C}_8\text{H}_{18}(l) + 25\text{O}_2(g) \rightarrow 16\text{CO}_2(g) + 18\text{H}_2\text{O}(g)
\]

### 2. **Determine the Standard Enthalpies of Formation:**

The enthalpy change (\(\Delta H\)) for the reaction can be calculated using the standard enthalpies of formation (\(\Delta H_f^\circ\)) of the reactants and products. The formula is:

\[
\Delta H_{reaction}^\circ = \sum \Delta H_f^\circ \text{(products)} - \sum \Delta H_f^\circ \text{(reactants)}
\]

The standard enthalpies of formation are typically given in tables:

- \(\Delta H_f^\circ\) for \( \text{C}_8\text{H}_{18}(l) \): -250.1 kJ/mol
- \(\Delta H_f^\circ\) for \( \text{O}_2(g) \): 0 kJ/mol (for elements in their standard state)
- \(\Delta H_f^\circ\) for \( \text{CO}_2(g) \): -393.5 kJ/mol
- \(\Delta H_f^\circ\) for \( \text{H}_2\text{O}(g) \): -241.8 kJ/mol

### 3. **Calculate the Enthalpy of Reaction:**

For the balanced equation:

\[
\Delta H_{reaction}^\circ = \left[16 \times (-393.5) + 18 \times (-241.8)\right] - \left[2 \times (-250.1) + 25 \times 0\right]
\]

Simplifying:

\[
\Delta H_{reaction}^\circ = \left[-6296 - 4352.4\right] - [-500.2]
\]

\[
\Delta H_{reaction}^\circ = -10648.4 + 500.2
\]

\[
\Delta H_{reaction}^\circ = -10148.2 \text{ kJ for 2 moles of octane}
\]

### 4. **Enthalpy for 1 Mole of Octane:**

Since the calculation was for 2 moles of octane, divide the result by 2:

\[
\Delta H_{reaction}^\circ = \frac{-10148.2}{2} \text{ kJ/mol} = -5074.1 \text{ kJ/mol}
\]

### **Final Answer:**

The enthalpy of reaction for the combustion of 1 mole of octane is **\(-5074.1 \, \text{kJ/mol}\)**.

---

Would you like a deeper explanation on any part of this process?

### Related Questions:
1. How do you interpret enthalpy values in exothermic and endothermic reactions?
2. What is the significance of the standard enthalpy of formation?
3. How do bond energies relate to enthalpy changes in chemical reactions?
4. How does the combustion of octane compare with other hydrocarbons?
5. How can enthalpy change be measured experimentally?

### Tip:
Always ensure the chemical equation is balanced before performing any thermodynamic calculations to avoid errors in your results.

Learn how to calculate the enthalpy of reaction for the combustion of octane using balanced chemical equations and standard enthalpies of formation. This problem involves balancing chemical equations and applying thermodynamic principles.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation