Question

Consider the following reaction: $$\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$ \Delta H=-891 \mathrm{kJ} $$ Calculate the enthalpy change for each of the following cases: a. 1.00 g methane is burned in excess oxygen. b. \(1.00 \times 10^{3}\) L methane gas at 740 . torr and \(25^{\circ} \mathrm{C}\) are burned in excess oxygen.

Short answer

The enthalpy change for burning 1.00 g of methane is -55.6 kJ, and the enthalpy change for burning \(1.00 \times 10^{3}\) L methane gas at 740 torr and 25ºC is -35,500 kJ.

Step-by-step solution

Calculate moles of methane burned in part a

Given mass of methane is 1.00 g. The molecular weight of methane (CH4) is 12.01 g/mol (C) + 4 * 1.01 g/mol (H) = 16.05 g/mol. We can calculate the moles of methane using the formula: Moles = (Mass)/(Molecular weight) Moles (CH4) = \( \frac{1.00}{16.05} \) = 0.0623 moles

Calculate enthalpy change for part a

The stoichiometry of the reaction is that 1 mole of CH4 produces 1 mole of CO2, and -891 kJ of heat is released. So for every mole of CH4 reacted, the enthalpy change is -891 kJ. We can use the ratio to determine the enthalpy change for 0.0623 moles of CH4 as follows: ΔH = -(891 kJ/mol) * (0.0623 moles) = -55.6 kJ Hence, the enthalpy change for burning 1.00 g of methane is -55.6 kJ. #b. \(1.00 \times 10^{3}\) L methane gas at 740 torr and \(25^{\circ} \mathrm{C}\) are burned in excess oxygen#

Calculate moles of methane burned in part b

In this case, we are given the volume, pressure, and temperature of the methane gas. We can use the ideal gas law equation (PV = nRT) to find the moles of CH4. Before we do that, we need to convert the given units to SI units: Pressure (P) = 740 torr * (101.325 kPa / 760 torr) = 98.5 kPa Temperature (T) = 25ºC + 273.15 = 298.15 K Volume (V) = 1.00 * 10^3 L Now, we can calculate the number of moles (n) using the ideal gas law equation: n = PV / RT R = 8.314 J/(mol K) n = \( \frac{98.5 \times 1000}{8.314 \times 298.15} \) = 39.9 moles

Calculate enthalpy change for part b

Similar to part a, we use the stoichiometry of the reaction and the enthalpy change to calculate the enthalpy change for burning 39.9 moles of CH4: ΔH = -(891 kJ/mol) * (39.9 moles) = -35,500 kJ Hence, the enthalpy change for burning \(1.00 \times 10^{3}\) L methane gas at 740 torr and 25ºC is -35,500 kJ.

Key concepts

Understanding Stoichiometry in Chemical Reactions

Stoichiometry is the calculation method that helps us understand the proportions of reactants and products in a chemical reaction. It is like a recipe that tells us how much of each ingredient is needed to get the desired result.
In the given reaction, one mole of methane (\( \text{CH}_4 \)) reacts with two moles of oxygen (\( \text{O}_2 \)) to produce one mole of carbon dioxide (\( \text{CO}_2 \)) and two moles of water (\( \text{H}_2\text{O} \)).
This tells us that if we have a different quantity of methane, we must adjust the amount of oxygen accordingly to maintain the balance of the equation. In chemical terms, this means using the molar ratios provided by the reaction's coefficients.
When dealing with enthalpy change, the stoichiometry helps us determine how much energy is released or absorbed when a certain amount of reactant is involved. For instance, burning one mole of methane releases 891 kJ of energy. If we only have 0.0623 moles as in part a, we adjust this value proportionately, hence the calculation of the enthalpy change.

Applying the Ideal Gas Law for Gas Calculations

The ideal gas law is a useful equation in chemistry that relates the four properties of a gas: pressure (\( P \)), volume (\( V \)), temperature (\( T \)), and moles (\( n \)). The formula is expressed as \( PV = nRT \).
In this exercise, we use the ideal gas law to determine the number of moles of methane (\( \text{CH}_4 \)) gas under specific conditions of pressure, temperature, and volume.

• Pressure: It was given in torr and should be converted to kilopascals (kPa) since the gas constant \( R \) we use is in kPa.
• Temperature: Convert Celsius to Kelvin by adding 273.15.
• Volume: Ensure that it is in liters.

Using these converted values in \( PV = nRT \) allows us to solve for \( n \), the moles of methane.
This step is essential because it allows us to further calculate the enthalpy change knowing exactly how many moles of gas we are working with.

Calculating Enthalpy Change in Chemical Reactions

Enthalpy change (\( \Delta H \)) is a key concept in thermodynamics that refers to the heat released or absorbed during a chemical reaction at constant pressure.
In the given exercise, the reaction of burning methane in oxygen results in a negative enthalpy change, meaning that it is exothermic—it releases energy.
For part a, we calculated the enthalpy change based on 0.0623 moles of methane. The stoichiometric information helps us proportionally adjust the heat released for the different moles involved.
For part b, after finding the moles using the ideal gas law, we used the reaction's enthalpy change per mole (\(-891 \text{kJ/mol}\)) in conjunction with the number of moles calculated from the gas conditions.
In both exercises, the steps ensure the proper application of concepts to find the respective enthalpy changes, illustrating how stoichiometry and gas laws come together to solve real-world problems in chemistry.