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Chapter 5: Problem 113

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 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 \(\Delta \mathrm{H}_{f}^{\circ}=-239 \mathrm{kJ} / \mathrm{mol}\)

Short Answer

Expert verified
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

01

Balanced Combustion Equations

First, let's write the balanced chemical equations for the combustion of ammonia and methanol: For ammonia: \(4 NH_3(g) + 3 O_2(g) \rightarrow 2 N_2(g) + 6 H_2O(g)\) For methanol: \(2 CH_3OH(l) + 3 O_2(g) \rightarrow 2 CO_2(g) + 4 H_2O(g)\)
02

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: \(\frac{0.81 \mathrm{g/cm^3} \times 1000 \mathrm{cm^3}}{17.03 \mathrm{g/mol}} = 47.56 \mathrm{mol}\) Moles of methanol in 1 L: \(\frac{0.792 \mathrm{g/cm^3} \times 1000 \mathrm{cm^3}}{32.04 \mathrm{g/mol}} = 24.72 \mathrm{mol}\)
03

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
04

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
05

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).

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Most popular questions from this chapter

When a 6.50 -g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee-cup calorimeter (Figure 5.18\()\) the temperature rises from 21.6 to to \(37.8^{\circ} \mathrm{C}\) . (a) Calculate the quantity of heat (in kJ) released in the reaction. (b) Using your result from part (a), calculate \(\Delta H\) (in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NaOH} )\) for the solution process. Assume that the specific heat of the solution is the same as that of pure water.

One of the best-selling light, or low-calorie, beers is 4.2\(\%\) alcohol by volume and a 12 -oz serving contains 110 Calories; remember: 1 Calorie \(=1000\) cal \(=1\) kcal. To estimate the percentage of Calories that comes from the alcohol, consider the following questions. (a) Write a balanced chemical equation for the reaction of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) , with oxygen to make carbon dioxide and water. (b) Use enthalpies of formation in Appendix \(\mathrm{C}\) to determine \(\Delta H\) for this reaction. \((\mathbf{c})\) If 4.2\(\%\) of the total volume is ethanol and the density of ethanol is \(0.789 \mathrm{g} / \mathrm{mL},\) what mass of ethanol does a 12 - oz serving of light beer contain? (\boldsymbol{d} ) How many Calories are released by the metabolism of ethanol, the reaction from part (a)? (e) What percentage of the 110 Calories comes from the ethanol?

Given the data $$\begin{aligned} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) & \Delta H=+180.7 \mathrm{kJ} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) & \Delta H=-113.1 \mathrm{kJ} \\ 2 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H=-163.2 \mathrm{kJ} \end{aligned}$$ use Hess's law to calculate \(\Delta H\) for the reaction $$\mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g)$$

Ozone, \(\mathrm{O}_{3}(g),\) is a form of elemental oxygen that plays an important role in the absorption of ultraviolet radiation in the stratosphere. It decomposes to \(\mathrm{O}_{2}(g)\) at room temperature and pressure according to the following reaction: $$2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g) \quad \Delta H=-284.6 \mathrm{kJ}$$ (a) What is the enthalpy change for this reaction per mole of \(\mathrm{O}_{3}(g) ?\) (b) Which has the higher enthalpy under these conditions, 2 \(\mathrm{O}_{3}(g)\) or 3 \(\mathrm{O}_{2}(g) ?\)

In a thermodynamic study, a scientist focuses on the properties of a solution in an apparatus as illustrated. A solution is continuously flowing into the apparatus at the top and out at the bottom, such that the amount of solution in the apparatus is constant with time. (a) Is the solution in the apparatus a closed system, open system, or isolated system? (b) If the inlet and outlet were closed, what type of system would it be?
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