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 \(25^{\circ} \mathrm{C}\) is burned in excess oxygen.

Short answer

The short answers for the given cases are as follows: a. The enthalpy change when \(1.00\,g\) of methane is burned in excess oxygen is \(-55.7\,kJ\). b. The enthalpy change when \(1.00 \times 10^{3}\,L\) of methane gas at \(740\,torr\) and \(25^{\circ}\mathrm{C}\) is burned in excess oxygen is \(-8.84 \times 10^{4}\,kJ\).

Step-by-step solution

(Given Information and Reaction)

(Write down the balanced chemical reaction: \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) And the given enthalpy change is \(\Delta H = -891\mathrm{~kJ/mol}\).)

(Case a: Steps)

(1. Convert grams of methane to moles. 2. Use stoichiometry to relate moles of methane to the enthalpy change. 3. Calculate the enthalpy change based on the moles of methane.)

(Case a: Converting grams to moles)

(The molar mass of methane is \(16.04 \mathrm{~g/mol}\). Therefore, we can determine moles of methane as follows: \[\mathrm{moles \ of \ CH_{4}} = \frac{1.00 \mathrm{\ g}}{16.04 \mathrm{~g/mol}}\])

(Case a: Stoichiometry and Enthalpy Change)

(Now, use stoichiometry with the balanced chemical equation and the given enthalpy change: \[\Delta H_{a} = (-891 \mathrm{~kJ/mol}) \times \frac{1.00 \mathrm{\ g}}{16.04 \mathrm{~g/mol}}\])

(Case a: Final Calculation)

(Plug in the values and calculate the enthalpy change: \[\Delta H_{a} = (-891\mathrm{~kJ}) \times \frac{1.00 \mathrm{\ g}}{16.04 \mathrm{~g/mol}} = -55.7 \mathrm{~kJ}\])

(Case b: Steps)

(1. Convert the given volume, pressure, and temperature of methane to moles using the ideal gas law. 2. Use stoichiometry to relate moles of methane to the enthalpy change. 3. Calculate the enthalpy change based on the moles of methane.)

(Case b: Converting Volume, Pressure, and Temperature to Moles)

(Using the ideal gas law, PV=nRT, we can find the moles of methane: \[\mathrm{n} = \frac{PV}{RT}\] Given \(V = 1.00 \times 10^3 \mathrm{~L}\), \(P = 740\,\text{Torr} = 740/760 \mathrm{~atm}\), \(T = 298 \mathrm{~K}\) and \(R = 0.0821 \mathrm{\ L \cdot atm/mol \cdot K}\), we can calculate the moles: \[\mathrm{n} = \frac{(740/760 \mathrm{~atm})(1.00 \times 10^3 \mathrm{~L})}{(0.0821 \mathrm{\ L\cdot atm/mol \cdot K})(298 \mathrm{~K})}\])

(Case b: Stoichiometry and Enthalpy Change)

(Now, use stoichiometry with the balanced chemical equation and the given enthalpy change: \[\Delta H_{b} = (-891 \mathrm{~kJ/mol}) \times \frac{(740/760 \mathrm{~atm})(1.00 \times 10^3 \mathrm{~L})}{(0.0821 \mathrm{\ L\cdot atm/mol \cdot K})(298 \mathrm{~K})}\])

(Case b: Final Calculation)

(Plug in the values and calculate the enthalpy change: \[\Delta H_{b} = (-891\mathrm{~kJ/mol}) \times \frac{(740/760 \mathrm{~atm})(1000 \mathrm{~L})}{(0.0821 \mathrm{\ L\cdot atm/mol \cdot K})(298 \mathrm{~K})} = -8.84 \times 10^4 \mathrm{~kJ}\])

Key concepts

Stoichiometry in Chemical Reactions

Stoichiometry is the aspect of chemistry that deals with determining the quantitative relationships between the reactants and products in a chemical reaction. It's akin to a recipe that outlines how much of each ingredient is needed to make a dish. In the context of the given exercise, stoichiometry is used to relate the amount of methane (CH₄) consumed to the amount of energy released (enthalpy change, eaH).

When calculating the enthalpy change for a certain amount of methane burned, the first step is to convert the mass or volume of methane into moles. Since stoichiometry is based on mole ratios, understanding the mole concept is crucial. Once we have the moles, we can use the mole ratio from the balanced chemical equation to calculate the energy change associated with the reaction, taking the stoichiometry into full account.

Ideal Gas Law and Its Application

The Ideal Gas Law is articulated as PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature in Kelvin. This law provides a mathematical relationship between these four variables for an ideal gas, which is a hypothetical gas that perfectly follows the kinetic molecular theory.

In practice, real gases at low pressure and high temperature can often be approximated as ideal gases. In the given problem, the Ideal Gas Law allows us to calculate the amount of methane in moles from its volume, pressure, and temperature, data points especially useful when the substance in question is a gas and its mass is not provided directly.

Energy Transformations in Chemical Reactions

Chemical reactions involve the breaking and forming of chemical bonds, a process that absorbs or releases energy. The enthalpy change (deltaH) of a reaction is a measure of the total heat absorbed or released when a reaction occurs at constant pressure. It is an expression of the energy changes accompanying a chemical reaction.

Energetically, exothermic reactions, such as the combustion of methane, release heat, resulting in a negative deltaH. This is observed in the given exercise where the enthalpy change for the combustion of methane is listed as -891 kJ/mol, indicating that the reaction releases energy to the surroundings. Understanding this concept is crucial for predicting the energy produced or consumed in chemical processes and for applying such knowledge to real-world problems such as fuel energy content calculations.