Question

Methyl tert-butyl ether (MTBE), \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}\), a gasoline additive used to boost octane ratings, has \(\Delta H_{\mathrm{f}}^{\circ}=-313.6 \mathrm{~kJ} / \mathrm{mol}\). Write a balanced equation for its combustion reaction, and calculate its standard heat of combustion in kilojoules.

Short answer

Balanced equation: \(2 \text{C}_5\text{H}_{12}\text{O} + 15 \text{O}_2 = 10 \text{CO}_2 + 12 \text{H}_2\text{O}\). Heat of combustion: \(-7273.8 \text{ kJ}\).

Step-by-step solution

Determine the combustion reaction

Combustion of MTBE involves its reaction with oxygen to produce carbon dioxide and water. The general equation for the combustion of an organic compound \((\text{C}_x\text{H}_y\text{O}_z)\) is: \[ \text{C}_x\text{H}_y\text{O}_z + m\,\text{O}_2 \rightarrow a\,\text{CO}_2 + b\,\text{H}_2\text{O} \]For MTBE \((\text{C}_5\text{H}_{12}\text{O})\), we initially write the unbalanced reaction: \[ \text{C}_5\text{H}_{12}\text{O} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]

Balance the combustion reaction

To balance the reaction, we need to ensure atoms are equal on both sides:- **Carbon balance:** There are 5 carbon atoms in MTBE, so we need 5 \(\text{CO}_2\) molecules.- **Hydrogen balance:** There are 12 hydrogen atoms in MTBE, so we need 6 \(\text{H}_2\text{O}\) molecules.- **Oxygen balance:** The right side has \((5 \times 2) + 6 = 16\) oxygen atoms. MTBE contributes 1 oxygen atom, so we need 7.5 \(\text{O}_2\) molecules for a total of \(15/2 = 7.5\) \(\text{O}_2\) molecules.The balanced equation is: \[ \text{C}_5\text{H}_{12}\text{O} + \frac{15}{2}\,\text{O}_2 \rightarrow 5\,\text{CO}_2 + 6\,\text{H}_2\text{O} \]

Convert to integer coefficients

To avoid dealing with fractions, multiply the entire equation by 2 to obtain integer coefficients:\[ 2 \text{C}_5\text{H}_{12}\text{O} + 15 \text{O}_2 \rightarrow 10 \text{CO}_2 + 12 \text{H}_2\text{O} \]

Calculate standard heat of combustion

The enthalpy change for the combustion (\( \Delta H^\circ_{ ext{comb}} \) is calculated using the formula: \[ \Delta H^\circ_{ ext{comb}} = n \times \Delta H_f^\circ \text{(reactants)} - m \times \Delta H_f^\circ \text{(products)} \]Given \( \Delta H_f^\circ \text{(MTBE)} = -313.6 \text{ kJ/mol} \), \( \Delta H_f^\circ \text{(CO}_2) = -393.5 \text{ kJ/mol} \), and \( \Delta H_f^\circ \text{(H}_2\text{O}) = -241.8 \text{ kJ/mol} \), the change of enthalpy reaction becomes: \[ 2(-313.6) + 0 \text{ (for } \text{O}_2) - (10 \times -393.5 + 12 \times -241.8) \]Thus resulting changes are: \[ \Delta H^\circ_{ ext{comb}} = -627.2 + (3935 + 2901.6) \text{ kJ} = -7273.8 \text{ kJ} \]

Final Result

The balanced equation for the combustion of MTBE is: \[ 2 \text{C}_5\text{H}_{12}\text{O} + 15 \text{O}_2 \rightarrow 10 \text{CO}_2 + 12 \text{H}_2\text{O} \]The standard heat of combustion is \(-7273.8 \text{ kJ}\).

Key concepts

Enthalpy Change

Enthalpy change is a key concept in chemistry that refers to the heat exchange during a reaction at constant pressure. When a substance undergoes a chemical reaction, it can either absorb or release energy, primarily as heat. This energy change is quantified in terms of enthalpy (denoted as \( \Delta H \)).
In combustion reactions, the change in enthalpy represents the energy released when a compound completely burns in oxygen. This energy release is why combustion reactions are typically exothermic — releasing heat.

• Exothermic: The reaction releases energy, \( \Delta H \) is negative.
• Endothermic: The reaction absorbs energy, \( \Delta H \) is positive.

In the context of MTBE's combustion, the enthalpy change is calculated using the standard enthalpies of formation for the reactants and products involved. The equation \[\Delta H^\circ_{comb} = \sum (\Delta H^\circ_f \text{(products)}) - \sum (\Delta H^\circ_f \text{(reactants)})\] helps determine the total energy change for converting reactants to products, illustrating the concept of energy conservation in chemical reactions.

Standard Heat of Combustion

The standard heat of combustion is a specific type of enthalpy change that measures the energy released when one mole of a substance burns completely in oxygen under standard conditions (298 K and 1 atm). It is denoted as \( \Delta H^\circ_{comb} \) and is usually expressed in kilojoules per mole (kJ/mol).
This value provides insights into the energy content of fuels, determining how much heat a substance will release upon burning.

• Negative \( \Delta H^\circ_{comb} \): indicates an exothermic reaction, typical for combustion.
• Positive \( \Delta H^\circ_{comb} \): indicates an endothermic reaction, although rare in combustion.

In practice, calculating this involves using the formula: \[\Delta H^\circ_{comb} = n \cdot \Delta H^\circ_f \text{(reactants)} - m \cdot \Delta H^\circ_f \text{(products)}\]Where:- \( n \) and \( m \) are stoichiometric coefficients from the balanced equation.- \( \Delta H^\circ_f \) refers to standard enthalpy of formation.
For MTBE, the calculation updates the enthalpy values of its reaction outcome, giving a comprehensive view of the energetics involved in its combustion.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures mass conservation across a reaction. In any chemical equation, the number of each type of atom on the reactant side must equal the number on the product side, abiding by the law of conservation of mass.
When balancing the combustion of MTBE, consider the following steps:

• Start with Carbon: Count the carbon atoms on both sides. MTBE has 5 carbon atoms, thus requiring 5 \( \text{CO}_2 \) molecules on the product side.
• Balance Hydrogen: 12 hydrogen atoms in MTBE translate to 6 \( \text{H}_2\text{O} \) molecules produced.
• Balance Oxygen: Calculate the total oxygen requirement from \( \text{CO}_2 \) and \( \text{H}_2\text{O} \) production, minus those already present in MTBE. Adjust \( \text{O}_2 \) accordingly.

Avoid fractional coefficients by multiplying through to obtain whole numbers. For example, multiplying the reaction by 2 finalizes it to integer coefficients, ensuring clarity and precision in chemical communication.