Question

Ammonia \(\left(\mathrm{NH}_{3}\right)\) boils at \(-33^{\circ} \mathrm{C} ;\) at this temperature it has a density of \(0.81 \mathrm{~g} / \mathrm{cm}^{3}\). The enthalpy of formation of \(\mathrm{NH}_{3}(g)\) is \(-46.2 \mathrm{~kJ} / \mathrm{mol}\), and the enthalpy of vaporization of \(\mathrm{NH}_{3}(l)\) is \(23.2 \mathrm{~kJ} / \mathrm{mol}\). Calculate the enthalpy change when \(1 \mathrm{~L}\) of liquid \(\mathrm{NH}_{3}\) is burned in air to give \(\mathrm{N}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\). How does this compare with \(\Delta H\) for the complete combustion of \(1 \mathrm{~L}\) of liquid methanol \(\mathrm{CH}_{3} \mathrm{OH}(l) ?\) For \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{l})\), the density at \(25^{\circ} \mathrm{C}\) is \(0.792 \mathrm{~g} / \mathrm{cm}^{3}\), and \(\Delta H_{f}^{\circ}\) equals \(-239 \mathrm{~kJ} / \mathrm{mol}\).

Short answer

The enthalpy change when 1 L of liquid ammonia is burned in air to give N2(g) and H2O(g) is -1097.25 kJ. Comparatively, the enthalpy change for the complete combustion of 1 L of liquid methanol is -5918.25 kJ, which is much greater than that of ammonia.

Step-by-step solution

Find moles of ammonia and methanol

First, find the amount of moles in 1 L of liquid ammonia and 1 L of liquid methanol using their given densities and molecular weights: Molecular weights: NH3 = 14 (N) + 3 * 1 (H) = 17 g/mol CH3OH = 12 (C) + 4 * 1 (H) + 16 (O) = 32 g/mol Ammonia: Density = 0.81 g/cm³ Since 1 L = 1000 cm³, mass of ammonia = density * volume = 0.81 g/cm³ * 1000 cm³ = 810 g Moles of ammonia = mass / molecular weight = 810 g / 17 g/mol = 47.65 mol Methanol: Density = 0.792 g/cm³ Mass of methanol = density * volume = 0.792 g/cm³ * 1000 cm³ = 792 g Moles of methanol = mass / molecular weight = 792 g / 32 g/mol = 24.75 mol

Write balanced chemical equations

Write the balanced chemical equations for the complete combustion of ammonia and methanol to give N2(g) and H2O(g): Ammonia: \(4 NH_{3}(g) + 3 O_{2}(g) \to 2 N_{2}(g) + 6 H_{2}O(g)\) Methanol: \(2 CH_{3}OH(g) + 3 O_{2}(g) \to 2 CO_{2}(g) + 4 H_{2}O(g)\)

Calculate enthalpy changes

To calculate the enthalpy changes for each process, first calculate the enthalpy change for the formation of ammonia and methanol using their respective enthalpies of formation. Then add the enthalpy of vaporization for ammonia (since we are given the value for ammonia in liquid form). Ammonia: Enthalpy of formation = -46.2 kJ/mol Enthalpy change in combustion = Enthalpy of formation * moles of ammonia ΔH_comb_ammonia = (-46.2 kJ/mol) * 47.65 mol = -2202.63 kJ Add enthalpy of vaporization for ammonia: ΔH_vaporization = 23.2 kJ/mol * 47.65 mol = +1105.38 kJ Enthalpy change when 1 L of liquid NH3 is burned = -2202.63 kJ + 1105.38 kJ = -1097.25 kJ Methanol: Enthalpy of formation = -239 kJ/mol Enthalpy change in combustion = Enthalpy of formation * moles of methanol ΔH_comb_methanol = (-239 kJ/mol) * 24.75 mol = -5918.25 kJ

Compare enthalpy changes

Now, we can compare the enthalpy changes for the complete combustion of 1 L of liquid ammonia and 1 L of liquid methanol: Enthalpy change for ammonia: -1097.25 kJ Enthalpy change for methanol: -5918.25 kJ The enthalpy change for the complete combustion of 1 L of liquid methanol is much greater than that of 1 L of liquid ammonia.

Key concepts

Enthalpy of Formation

Enthalpy of formation is a fundamental concept in thermochemistry. It is the enthalpy change that occurs when one mole of a compound is formed from its elements in their standard states. This value is crucial for understanding the energy involved in creating one mole of a substance from its components.For example, in the given exercise, the enthalpy of formation of gaseous ammonia \((NH_3(g))\) is -46.2 kJ/mol. A negative value indicates that the formation of ammonia from nitrogen and hydrogen is an exothermic process, meaning it releases energy to the surroundings. Conversely, a positive enthalpy of formation would denote an endothermic reaction, where energy is absorbed.Understanding the enthalpy of formation helps predict reaction behavior and stability of the substances. It is also central in calculating other thermodynamic properties, such as the enthalpy change of a reaction, by using Hess's Law or through combining multiples of formation reactions.

Enthalpy of Vaporization

The enthalpy of vaporization tells us the energy required to vaporize a mole of liquid at its boiling point. It measures the strength of intermolecular forces in a liquid. A higher enthalpy of vaporization means stronger forces are present and more energy is needed for the phase change.For ammonia \((NH_3)\), the enthalpy of vaporization is 23.2 kJ/mol. This value shows the energy needed to convert liquid ammonia into gaseous ammonia at its boiling point, \(-33^\circ C\). When calculating the total enthalpy change in a reaction involving a phase change from liquid to gas, it's important to add this energy requirement to the overall process.This is crucial in real-world applications, such as industrial and refrigeration processes, where vapors are created as part of the energy cycle. Here, the enthalpy of vaporization must be considered to accurately manage energy efficiency and costs.

Chemical Equations

A chemical equation is a representation of a chemical reaction, showing the reactants and products, their physical states, and the stoichiometrical coefficients indicating the ratios in which substances react and are created.In the exercise, balanced chemical equations are provided for the combustion of ammonia and methanol. Using stoichiometry, the equations ensure that matter is conserved. For ammonia, the balanced equation is \(4NH_3(g) + 3O_2(g) \rightarrow 2N_2(g) + 6H_2O(g)\). For methanol, it's \(2CH_3OH(g) + 3O_2(g) \rightarrow 2CO_2(g) + 4H_2O(g)\).Balancing chemical equations correctly is vital for accurately calculating enthalpy changes, predicting yields, and understanding the proportion of reactants needed. With balanced equations, one can calculate the moles of reactants required and products formed, which is essential for determining energy changes in reactions.

Combustion Reactions

Combustion reactions involve the burning of a substance in the presence of oxygen, producing heat and light. These reactions are typically exothermic, releasing energy to the surroundings. They usually involve hydrocarbons or similar molecules reacting with oxygen to produce carbon dioxide and water.In the case of ammonia combustion \(4NH_3(g) + 3O_2(g) \rightarrow 2N_2(g) + 6H_2O(g)\), the reaction releases energy, indicated by the negative enthalpy change. Likewise, methanol combusts to form carbon dioxide and water, also releasing a significant amount of energy \(\Delta H = -5918.25\ kJ\) for 1 L.Understanding combustion is crucial in energy chemistry and for applications ranging from internal combustion engines to industrial power generation. Knowing how much energy a fuel releases and comparing different fuels' enthalpy changes can inform choices about the most efficient and cost-effective energy sources.