Question

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

Short answer

The enthalpy changes for burning 1.00 g of methane in excess oxygen and 1.00 x 10³ L methane gas at 740 torr and 25°C in excess oxygen are -55.6 kJ and -35,500 kJ, respectively.

Step-by-step solution

a. Calculate the enthalpy change when 1.00 g methane is burned in excess oxygen

To calculate the enthalpy change when 1.00 g of methane is burned in excess oxygen, first, we need to find the number of moles of methane (n) using its molar mass: 1.00 g CH4 * \(\frac{1 \text{ mol CH4}}{16.04 \text{ g CH4}}\)= 0.0624 mol CH4 Now, using the stoichiometry of the balanced chemical equation and the standard enthalpy change for the reaction, we can calculate the enthalpy change for the reaction: Enthalpy change = \(\Delta H_{rxn}\times\frac{mol\; of\; CH_{4}}{mol\;reaction}\) Enthalpy change =(-891 kJ/mol) × (0.0624 mol CH4) Enthalpy change = -55.6 kJ

b. Calculate the enthalpy change when 1.00 x 10³ L methane gas at 740 torr and 25°C are burned in excess oxygen

First, we need to convert the given pressure from torr to atm: P(atm) = \(\frac{740 \text{ torr}}{760 \text{ torr/atm}}\) = 0.974 atm Next, we need to convert the given temperature to Kelvin: T(K) = 25°C + 273.15 = 298.15 K Now, we can use the Ideal Gas Law (PV = nRT) to find the number of moles of methane: n = \(\frac{PV}{RT}\) n = \(\frac{(0.974 \text{ atm})(1.00 \times 10^{3} \text{ L})}{(0.0821 \frac{\text{ L atm}}{\text{K mol}})(298.15 \text{ K})}\) n = 39.86 mol CH4 Now, using the stoichiometry of the balanced chemical equation and the standard enthalpy change for the reaction, we can calculate the enthalpy change for the reaction: Enthalpy change = \(\Delta H_{rxn}\times\frac{mol\; of\; CH_{4}}{mol\;reaction}\) Enthalpy change = (-891 kJ/mol) × (39.86 mol CH4) Enthalpy change = -35,500 kJ The enthalpy changes for burning 1.00 g of methane in excess oxygen and 1.00 x 10³ L methane gas at 740 torr and 25°C in excess oxygen are -55.6 kJ and -35,500 kJ, respectively.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It allows us to understand the relationships between different substances involved in a reaction. For the given reaction \[ \mathrm{CH}_{4}(g) + 2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(l) \]we can see that one mole of methane reacts with two moles of oxygen. This balanced equation is crucial because it defines the ratio in which the reactants combine and the products form.
To calculate the enthalpy change when burning methane, we first need to determine how many moles of methane are being used. In the exercise, we start by converting grams of methane to moles. By using the molar mass of methane (16.04 g/mol), we can transform mass measurements into a format suitable for stoichiometric calculations. This makes it possible to apply the given enthalpy change per mole to the specific amount of substance we are examining.

Ideal Gas Law

The ideal gas law is a well-known equation that describes the behavior of gases. It is expressed as:\[ PV = nRT \]where:

• \(P\) stands for pressure
• \(V\) is volume
• \(n\) is the number of moles
• \(R\) is the ideal gas constant
• \(T\) is temperature in Kelvin

In our problem, we use this equation to find out how many moles of methane gas we have in a given volume under specified conditions (pressure and temperature). By converting pressure from torr to atm and converting temperature from Celsius to Kelvin, we ensure all units are compatible with the ideal gas constant, \(R = 0.0821 \text{ L atm/K mol}\).
This conversion enables us to solve the equation for \(n\), giving us the amount of gas involved, which is essential in determining the enthalpy change for a given reaction scenario.

Enthalpy Change

Enthalpy change, \(\Delta H\), is the measure of heat change at constant pressure. In chemical reactions, it tells us how much energy is absorbed or released as the reaction progresses. For our example reaction:\[ \mathrm{CH}_{4}(g) + 2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) + 2 \mathrm{H}_{2} \mathrm{O}(l) \]the given enthalpy change is \(-891 \text{ kJ/mol}\), indicating that the reaction is exothermic (energy is released).
Using the stoichiometry of the reaction, we calculate the enthalpy change for a specific amount of methane. This involves multiplying the standard enthalpy change by the number of moles of methane calculated earlier. This process shows how much total energy is released when methane is burned in each scenario presented in the exercise.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In this exercise, the reaction of methane with oxygen exemplifies a combustion reaction, where the reactants are oxidized to produce carbon dioxide and water, releasing energy in the form of heat.
Understanding the specifics of this reaction type allows us to predict the products and comprehend the energy changes involved. In combustion reactions, fuel consumption (methane in this case) leads to a consistent release of energy, which can be measured and calculated using tools like stoichiometry and enthalpy change calculations. Recognizing these transformation patterns is key to mastering chemical reaction dynamics.