Problem 138 - General Chemistry

Question

A \(10.0-\mathrm{g}\) sample of a mixture of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{4}\) reacts with oxygen at \(25^{\circ} \mathrm{C}\) and 1 atm to produce \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) If the reaction produces \(520 \mathrm{~kJ}\) of heat, what is the mass percentage of \(\mathrm{CH}_{4}\) in the mixture?

Step-by-Step Solution

Answer

The mass percentage of \( \mathrm{CH}_4 \) in the mixture is approximately 45.1\%.}

Step 1: Write the Balanced Chemical Equations

Firstly, determine the balanced chemical equations for the combustion of \( \mathrm{CH}_4 \) and \( \mathrm{C}_2 \mathrm{H}_4 \). For \( \mathrm{CH}_4 \): \( \mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \). For \( \mathrm{C}_2\mathrm{H}_4 \): \( \mathrm{C}_2\mathrm{H}_4 + 3\mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \).

Step 2: Heat of Reaction for Each Component

Determine the heat of combustion for each component: \(\Delta H_{\mathrm{CH}_4} = -890.4 \text{ kJ/mol} \) and \(\Delta H_{\mathrm{C}_2\mathrm{H}_4} = -1411.1 \text{ kJ/mol} \). These values represent the heat released per mole during combustion.

Step 3: Set Up Equations for Mass and Heat

Let \( m_{\mathrm{CH}_4} \) and \( m_{\mathrm{C}_2\mathrm{H}_4} \) be the masses of \( \mathrm{CH}_4 \) and \( \mathrm{C}_2\mathrm{H}_4 \) respectively. We have the equations: \( m_{\mathrm{CH}_4} + m_{\mathrm{C}_2\mathrm{H}_4} = 10 \) g. We also know that the reaction heat produced is the sum from both components: \( \frac{m_{\mathrm{CH}_4}}{16.0} \times 890.4 + \frac{m_{\mathrm{C}_2\mathrm{H}_4}}{28.0} \times 1411.1 = 520 \text{ kJ} \).

Step 4: Solve the System of Equations

Substitute \( m_{\mathrm{C}_2\mathrm{H}_4} = 10 - m_{\mathrm{CH}_4} \) into the heat equation and solve:\[ \frac{m_{\mathrm{CH}_4}}{16.0} \times 890.4 + \frac{10 - m_{\mathrm{CH}_4}}{28.0} \times 1411.1 = 520 \]Simplify and solve this equation to find \( m_{\mathrm{CH}_4} \approx 4.51 \text{ g} \).

Step 5: Calculate Mass Percentage of \( \mathrm{CH}_4 \)

The mass percentage of \( \mathrm{CH}_4 \) is calculated as:\[ \text{Mass \, percentage} = \left( \frac{m_{\mathrm{CH}_4}}{10.0} \right) \times 100\% \approx 45.1\% \]

Key Concepts

Chemical Reactions

Chemical reactions are processes in which substances undergo transformation by breaking and forming new bonds, resulting in new substances. These are described by chemical equations, which must be balanced to reflect the conservation of mass. For example, in our exercise, methane (\( \mathrm{CH}_4 \)) and ethene (\( \mathrm{C}_2 \mathrm{H}_4 \)) react with oxygen to form carbon dioxide and water. It is crucial to write balanced chemical equations to determine the stoichiometry of the reactants and products involved:

• Methane combustion: \( \mathrm{CH}_4 + 2\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \)
• Ethene combustion: \( \mathrm{C}_2\mathrm{H}_4 + 3\mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \)

Balanced equations ensure that there is an equal number of each type of atom on both sides, which is fundamental for performing stoichiometric calculations. This balance allows us to keep track of how much of each reactant is used and how much of each product is formed, guiding us through quantitative analyses in chemistry.

Combustion

Combustion is a specific type of chemical reaction that occurs when a substance reacts with oxygen, releasing energy in the form of heat and light. It involves hydrocarbons like methane and ethene, which, when burned in the presence of oxygen, produce carbon dioxide and water. Combustion reactions are exothermic, meaning they release energy. Here's how it works in the context of the exercise:

• When methane \( \mathrm{CH}_4 \) combusts, it reacts with oxygen \( \mathrm{O}_2 \), forming carbon dioxide \( \mathrm{CO}_2 \) and water \( \mathrm{H}_2\mathrm{O} \), releasing 890.4 kJ/mol of energy.
• Ethene \( \mathrm{C}_2 \mathrm{H}_4 \) combusts similarly, releasing 1411.1 kJ/mol of energy.

The energy released in these reactions is crucial for calculating how much heat is produced in an entire process or determining the energy efficiency of fuels. Understanding combustion helps us grasp the broader impacts of fuel use, including energy generation and environmental effects.

Enthalpy Change

Enthalpy change (\( \Delta H \)) is the measure of heat change during a chemical reaction, conducted at constant pressure. During reactions like combustion, the enthalpy change indicates whether energy is absorbed or released. For the exercise, let's delve into why this is important:

• For combustion of methane, enthalpy change \( \Delta H_{\mathrm{CH}_4} = -890.4 \text{ kJ/mol} \) signifies an exothermic process due to negative value.
• Similarly, for ethene, \( \Delta H_{\mathrm{C}_2\mathrm{H}_4} = -1411.1 \text{ kJ/mol} \).

The negative sign shows that heat is being released as the reaction advances, integral for determining how much of each substance combusted. To solve for the mass percentage of methane in the mixture, this enthalpy data helped calculate how much energy comes from each component. Thus, understanding enthalpy changes aids in comparative studies of fuel efficiencies and environmental impacts, supporting sustainability efforts.