Problem 25 - General Chemistry: Principles and Modern Applications

Question

How many liters of \(\mathrm{CH}_{4}(\mathrm{g}),\) measured at \(23.4^{\circ} \mathrm{C}\) and \(768 \mathrm{mmHg},\) must be burned to provide the heat needed to vaporize 3.78 L of water at \(100^{\circ} \mathrm{C}\) ? \(\Delta \mathrm{H}_{\text {combustion }}=\) \(-8.90 \times 10^{2} \mathrm{kJmol}^{-1} \mathrm{CH}_{4} \quad\) For \(\quad \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad\) at \(\quad 100^{\circ} \mathrm{C}\) \(d=0.958 \mathrm{g} \mathrm{cm}^{-3},\) and \(\Delta H_{\mathrm{vap}}=40.7 \mathrm{kJmol}^{-1}\)

Step-by-Step Solution

Answer

So, about 225 liters of \(\mathrm{CH}_{4}(\mathrm{g})\) are needed to vaporize 3.78 liters of water at \(100^{\circ} \mathrm{C}\).

Step 1: Calculate the moles of water

We start by finding the number of moles of water that we have. Since \(1 \mathrm{cm}^{3}=1 \mathrm{mL}\), and \(1 \mathrm{L}=1000 \mathrm{mL}\), we can convert the volume of water in liters to volume in cubic centimeters. Hence, 3.78 L of water equals 3.78 × 1000 = 3780 cm3 (or mL). From the density value given, 0.958 g of water has a volume of 1 cm3, so 3780 cm3 of water has mass = 0.958 g/cm3 x 3780 cm3 = 3621.64 g. We then use the molar mass of water (18.02 g/mol) to find the number of moles of water, which equals 3621.64 g ÷ 18.02 g/mol = 200.97 moles of water.

Step 2: Calculate the heat for vaporization

To calculate the required heat for vaporization, we use the formula q = nΔH where q represents the heat, n the number of moles, and ΔH the heat of vaporization. Plugging in the values, the heat required = 200.97 moles x 40.7 kJ/mol = 8179.5 kJ.

Step 3: Calculate the moles of methane

We want to find how much CH4 should be burned to provide this heat. Since the enthalpy of combustion ΔH = -890 kJ/mol, this means that burning 1 mol of CH4 generates 890 kJ. Using the formula q = nΔH where q represents the heat and η the number of moles, the number of moles of CH4 needed equals heat required ÷ absolute value of enthalpy of combustion = 8179.5 kJ ÷ 890 kJ/mol = 9.19 moles of CH4.

Step 4: Calculate the volume of methane

To convert the number of moles of methane to volume, we can use the ideal gas law, PV=nRT. Rearranging to solve for V gives V = nRT/P. Given that R = 0.08206 L.atm/(mol.K), the temperature T = 23.4 + 273.15 = 296.55 K (converting Celsius to Kelvin), and the pressure P = 768 mmHg = 1.01 atm (converting mmHg to atm), the volume required = 9.19 mol x 0.08206 L.atm/(mol.K) x 296.55 K ÷ 1.01 atm = 224.7 L.

Key Concepts

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert a specific amount of a substance from a liquid to a gas at constant temperature and pressure. When water vaporizes, it needs a significant amount of energy due to the hydrogen bonds between its molecules.
The heat of vaporization is denoted as \( \Delta H_{\text{vap}} \), and it's usually expressed in kilojoules per mole (kJ/mol). For water, at its boiling point of 100°C, the heat of vaporization is 40.7 kJ/mol. This indicates that 40.7 kJ of energy is needed to vaporize one mole of water.

• This property is critical in many applications, such as in steam engines, cooking, and even in climate processes like evaporation.
• To calculate the total energy needed to vaporize a given amount of water, multiply the number of moles by the heat of vaporization: \( q = n \Delta H_{\text{vap}} \), where \ q \ is the heat energy needed, and \ n \ is the moles of water.

Understanding this concept helps us calculate how much energy is needed to transform water from its liquid state to a gaseous state.

Combustion Enthalpy

Combustion enthalpy is the amount of energy released when a substance undergoes complete combustion with oxygen. In the case of methane (CH_4), the combustion process is highly exothermic, meaning it releases a lot of energy.
The enthalpy of combustion is represented as \( \Delta H_{\text{combustion}} \), and for methane, it's given as -890 kJ/mol. This negative sign indicates that energy is released, and 890 kJ are produced per mole of methane combusted.

• Enthalpy values like this help us understand energy changes during reactions and are important for designing energy-efficient processes.
• The formula to calculate the energy released or consumed is \ q = n \Delta H_{\text{combustion}} \, where \ q \ is the heat change, and \ n \ is the number of moles of the substance.
• In practical applications, this concept is different from theoretical limits because factors like incomplete combustion or side reactions might occur.

Understanding combustion enthalpy allows us to calculate how much fuel is needed to produce a specific quantity of energy, such as that required for vaporizing a certain amount of water.

Methane Combustion

Methane combustion is the reaction where methane (CH_4) burns in the presence of oxygen to produce carbon dioxide, water, and energy. This reaction is essential in many industrial processes and is a primary source of energy in gas stoves, boilers, and power plants.
The balanced chemical equation for methane combustion is:\[ \text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g) + \text{energy} \]

• Methane combustion is vital due to the efficiency and cleanliness of the reaction, primarily producing water and carbon dioxide.
• This process is used for heating, generating electricity, and even fuel for transportation.
• The simplicity of methane's molecular structure (one carbon atom bonded to four hydrogen atoms) allows it to combust cleanly, reducing unwanted by-products.

In our calculations, knowing how much methane to burn helps us supply the exact energy needed, like in the example of vaporizing water. The efficiency of methane makes it a crucial element in our energy landscape.