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 $\mathrm{\Delta }H$ for the complete combustion of 1 Lof liquid methanol, ${\mathrm{CH}}_3\mathrm{OH}(l)?$ For ${\mathrm{CH}}_3\mathrm{OH}(l),$ the density at ${25}^{\circ }\mathrm{C}$ is $0.792\mathrm{g}/{\mathrm{cm}}^3,$ and $\mathrm{\Delta }{\mathrm{H}}_f^{\circ }=-239\mathrm{kJ}/\mathrm{mol}$

Short answer

The enthalpy change for the combustion of 1 L of liquid ammonia is -356.568 kJ, while the enthalpy change for the combustion of 1 L of liquid methanol is 4,153.48 kJ. This comparison shows that the combustion of 1 L of methanol releases more energy than the combustion of 1 L of liquid ammonia.

Step-by-step solution

Balanced Combustion Equations

First, let's write the balanced chemical equations for the combustion of ammonia and methanol: For ammonia: $4N{H}_3(g)+3O_2(g)\to 2N_2(g)+6H_{2}O(g)$ For methanol: $2C{H}_{3}OH(l)+3O_2(g)\to 2C{O}_2(g)+4H_{2}O(g)$

Determine the Moles in 1L

Next, we need to calculate the number of moles in 1 L of liquid ammonia and liquid methanol. To do this, we will use the densities given in the problem. Moles of ammonia in 1 L: $\frac{0.81\mathrm{g}/{\mathrm{cm}}^3\times 1000{\mathrm{cm}}^3}{17.03\mathrm{g}/\mathrm{mol}}=47.56\mathrm{mol}$ Moles of methanol in 1 L: $\frac{0.792\mathrm{g}/{\mathrm{cm}}^3\times 1000{\mathrm{cm}}^3}{32.04\mathrm{g}/\mathrm{mol}}=24.72\mathrm{mol}$

Calculate Enthalpy Change for Ammonia Combustion

Now we can calculate the enthalpy change for the combustion of 1 L of ammonia. First, we need to find the total enthalpy change due to the formation of 6 moles of water vapor: Enthalpy change = (moles of water vapor produced) × (enthalpy of formation of water vapor) = (6 moles) × (-241.8 kJ/mol) = -1450.8 kJ Next, the enthalpy change due to vaporization of liquid ammonia: Enthalpy change = (moles of ammonia) × (enthalpy of vaporization of ammonia) = (47.56 moles) × (23.2 kJ/mol) = 1103.36 kJ Finally, the enthalpy change due to the formation of ammonia gas: Enthalpy change = (moles of ammonia) × (enthalpy of formation of ammonia gas) = (47.56 moles) × (-46.2 kJ/mol) = -2197.592 kJ Now, we can find the total enthalpy change in the combustion of 1 L of ammonia: Total enthalpy change = (-1450.8 - 1103.36 + 2197.592) kJ = -356.568 kJ

Calculate Enthalpy Change for Methanol Combustion

Next, we can calculate the enthalpy change for the combustion of 1 L of methanol. Enthalpy change due to formation of 4 moles of water vapor: Enthalpy change = (moles of water vapor) × (enthalpy of formation of water vapor) = (4 moles) × (-241.8 kJ/mol) = -967.2 kJ Enthalpy change due to the formation of 2 moles of CO2: Enthalpy change = (moles of CO2) × (enthalpy of formation of CO2) = (2 moles) × (-393.5 kJ/mol) = -787 kJ Finally, the enthalpy change due to the formation of methanol gas: Enthalpy change = (moles of methanol) × (enthalpy of formation of methanol gas) = (24.72 moles) × (-239 kJ/mol) = -5907.68 kJ Now, we can find the total enthalpy change in the combustion of 1 L of methanol: Total enthalpy change = (-967.2 - 787 + 5907.68) kJ = 4153.48 kJ

Compare the Enthalpy Changes

Finally, we can compare the enthalpy changes for the combustion of 1 L of ammonia and 1 L of methanol: Enthalpy change for ammonia combustion: -356.568 kJ Enthalpy change for methanol combustion: 4153.48 kJ From these calculations, we can conclude that the combustion of 1 L of methanol releases more energy (4,153.48 kJ) than the combustion of 1 L of liquid ammonia (-356.568 kJ).

Key concepts

enthalpy of formation

The enthalpy of formation is a measure of the energy change when one mole of a compound is formed from its constituent elements under standard conditions. This value can be positive or negative and provides insight into the stability of compounds.

For example, ammonia (${\mathrm{NH}}_3\left(\text{g}\right)$) has a negative enthalpy of formation of $-46.2\mathrm{kJ}/\mathrm{mol}$. This indicates that energy is released when ammonia is formed from nitrogen and hydrogen gases, suggesting stability in the product.

Understanding this concept helps predict reaction outcomes and energy requirements. It's a fundamental aspect when considering reactions involving heat changes, such as combustion.

enthalpy of vaporization

Enthalpy of vaporization is the energy required to convert a substance from a liquid to a gas at its boiling point. Often expressed in $\mathrm{kJ}/\mathrm{mol}$, it reflects the energy needed to overcome intermolecular forces.

In the case of liquid ammonia, the enthalpy of vaporization is $23.2\mathrm{kJ}/\mathrm{mol}$. This value touches upon the strength of interactions in liquid ammonia that must be overcome to vaporize the substance.

The concept is crucial in processes like boiling, distillation, and understanding the energetics of phase changes. High enthalpies of vaporization mean stronger intermolecular attractions and higher energy required for vaporization.

combustion reaction

A combustion reaction is a chemical reaction where a substance reacts with oxygen to produce heat, often yielding gases like ${\mathrm{CO}}_2$ and ${\mathrm{H}}_2\mathrm{O}$. These reactions are typically exothermic, releasing a significant amount of energy.

The combustion of ammonia and methanol are examples noted here, producing different products:

• Ammonia combustion: yields ${\mathrm{N}}_2$ and ${\mathrm{H}}_2\mathrm{O}$.
• Methanol combustion: results in ${\mathrm{CO}}_2$ and ${\mathrm{H}}_2\mathrm{O}$.

In practice, combustion reactions are employed for energy generation, from engines to power plants. Analyzing these reactions involves enthalpy calculations to determine energy output efficiency.

mole calculations

Mole calculations involve determining the number of moles of a substance in a given volume or mass. They are essential for stoichiometric calculations in chemistry.

Relevant to the exercise, we calculated moles by using densities:

• Ammonia: $\frac{0.81\mathrm{g}/{\mathrm{cm}}^3\times 1000{\mathrm{cm}}^3}{17.03\mathrm{g}/\mathrm{mol}}=47.56\mathrm{mol}$
• Methanol: $\frac{0.792\mathrm{g}/{\mathrm{cm}}^3\times 1000{\mathrm{cm}}^3}{32.04\mathrm{g}/\mathrm{mol}}=24.72\mathrm{mol}$

Mole calculations serve as a backbone for more complex calculations like determining the enthalpy change of reactions or the yield of chemical processes. They relate macroscopic measurements like mass and volume to the microscopic scale of molecules.