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When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates $$\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s) \quad \Delta H=-65.5 \mathrm{kJ}$$ (a) Calculate \(\Delta H\) for the production of 0.450 mol of AgCl by this reaction. (b) Calculate \(\Delta H\) for the production of 9.00 \(\mathrm{g}\) of AgCl. (c) Calculate \(\Delta H\) when \(9.25 \times 10^{-4} \mathrm{mol}\) of AgCl dissolves in water.

Short Answer

Expert verified
\( \Delta H_{0.450} = -29.475\,\mathrm{kJ} \) \( \Delta H_{9.00} = -4.110\,\mathrm{kJ} \) \( \Delta H_{9.25 \times 10^{-4}} = +0.06058\,\mathrm{kJ} \)

Step by step solution

01

To calculate 螖H for 0.450 mol of AgCl production, we start with 螖H per mole (given in the exercise) and multiply it by the number of moles (0.450 mol): $$ \Delta H_{0.450} = \Delta H \times n $$ $$ \Delta H_{0.450} = -65.5\,\mathrm{kJ/mol} \times 0.450\,\mathrm{mol} $$tag_title# Calculate the 螖H value for 0.450 mol of AgCl formed.

Now, we can calculate the 螖H value for 0.450 mol of AgCl formed: $$ \Delta H_{0.450} = -29.475\,\mathrm{kJ} $$ #b) 螖H for 9.00 g of AgCl production:
02

Calculate the moles of AgCl from the mass given.

First, we need to convert the mass of AgCl (9.00 g) into moles. To do this, we use the molar mass of silver chloride (AgCl): $$ \mathrm{Molar\,mass\,of\, AgCl} = M_{\mathrm{Ag}} + M_{\mathrm{Cl}} = 107.87\,\mathrm{g/mol} + 35.45\,\mathrm{g/mol} = 143.32\,\mathrm{g/mol} $$ $$ n = \frac{m}{M} = \frac{9.00\,\mathrm{g}}{143.32\,\mathrm{g/mol}} $$ Now we calculate the moles of AgCl: $$ n = 0.06277\,\mathrm{mol} $$tag_title# Calculate the 螖H value for 9.00 g of AgCl formed.
03

Now, we can calculate the 螖H value for 9.00 g (0.06277 mol) of Ag Cl formed: $$ \Delta H_{9.00} = \Delta H \times n $$ $$ \Delta H_{9.00} = -65.5\,\mathrm{kJ/mol} \times 0.06277\,\mathrm{mol} $$ Now we calculate the 螖H value for 9.00 g (0.06277 mol) of AgCl formed: $$ \Delta H_{9.00} = -4.110\,\mathrm{kJ} $$ #c) 螖H for 9.25 x 10鈦烩伌 mol of AgCl dissolution:

For the reverse reaction, change the sign of 螖H.
04

Since dissolution is the reverse reaction, we need to change the sign of 螖H: $$ \Delta H_{dissolution} = +65.5\,\mathrm{kJ/mol} $$tag_title# Calculate the 螖H value for 9.25 x 10鈦烩伌 mol of AgCl dissolved.

Now, we calculate the 螖H value for 9.25 x 10鈦烩伌 mol of AgCl dissolved: $$ \Delta H_{9.25 \times 10^{-4}} = \Delta H_{dissolution} \times n $$ $$ \Delta H_{9.25 \times 10^{-4}} = +65.5\,\mathrm{kJ/mol} \times 9.25 \times 10^{-4}\,\mathrm{mol} $$ Now we calculate the 螖H value for 9.25 x 10鈦烩伌 mol of AgCl dissolved: $$ \Delta H_{9.25 \times 10^{-4}} = +0.06058\,\mathrm{kJ} $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Thermodynamics
Chemical thermodynamics is a fascinating study that combines the principles of physics and chemistry to analyze the energy changes associated with chemical reactions and processes. In the context of our problem, we focus on a particular aspect of thermodynamics called enthalpy (represented as \( \Delta H \)), which expresses the heat change at constant pressure. \( \Delta H \) is essential for predicting whether reactions will occur spontaneously and for understanding how much energy is released or absorbed during a reaction.

Enthalpy change is measured in kilojoules per mole (kJ/mol) and can be positive (endothermic reaction) or negative (exothermic reaction). Here, our textbook problem indicates that the formation of silver chloride (AgCl) from its ions is exothermic, as \( \Delta H \) is a negative value (-65.5 kJ/mol). This negative sign signifies that the reaction releases energy to the surroundings, and is thermodynamically favorable under standard conditions.
Precipitation Reactions
Precipitation reactions are a type of chemical reaction that occur when two soluble salts in aqueous solutions are mixed together to form an insoluble solid, known as a precipitate. In our textbook exercise, silver ions \( (\mathrm{Ag}^{+}) \) and chloride ions \( (\mathrm{Cl}^{-}) \) in solution combine to form silver chloride \( (\mathrm{AgCl}) \) as a solid precipitate.

These reactions are central to various fields such as analytical chemistry, environmental science, and materials science. Understanding how to calculate the enthalpy change, \( \Delta H \), for precipitation reactions allows students to predict the energy required or released when forming a certain amount of precipitate. This can aid in the design and optimization of chemical processes, including the purification of substances or the manufacture of advanced materials.
Mole Concept
The mole concept is a foundational principle in chemistry that enables chemists to count particles like atoms, molecules, or ions by weighing them. One mole corresponds to Avogadro's number (approximately \(6.022 \times 10^{23}\) entities) and is defined as the quantity of a substance that contains as many particles as there are atoms in 12 grams of carbon-12.

Calculations involving moles are crucial when dealing with chemical equations and reactions. In the exercise, for instance, to calculate the enthalpy changes for different amounts of AgCl, we convert grams to moles using the molar mass of AgCl, then multiply by the enthalpy change per mole. This application of the mole concept enables precise predictions about the energy changes that come with the formation or dissolution of compounds, highlighting its importance in chemical thermodynamics and quantitative analysis.

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Most popular questions from this chapter

A 1.800 -g sample of phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) was burned in a bomb calorimeter whose total heat capacity is 11.66 \(\mathrm{kJ} /^{\circ} \mathrm{C}\) The temperature of the calorimeter plus contents increased from 21.36 to \(26.37^{\circ} \mathrm{C}\) (a) Write a balanced chemical equation for the bomb calorimeter reaction. (b) What is the heat of combustion per gram of phenol? Per mole of phenol?

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(a) What is meant by the term fuel value? (b) Which is a greater source of energy as food, 5 g of fat or 9 g of carbohydrate? (c) The metabolism of glucose produces \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) How does the human body expel these reaction products?

The Sun supplies about 1.0 kilowatt of energy for each square meter of surface area \(\left(1.0 \mathrm{kW} / \mathrm{m}^{2},\) where a watt \(=1 \mathrm{J} / \mathrm{s}\right)\) Plants produce the equivalent of about 0.20 \(\mathrm{g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per square meter. Assuming that the sucrose is produced as follows, calculate the percentage of sunlight used to produce sucrose. $$\begin{array}{c}{12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2}(g)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {\Delta H=5645 \mathrm{kJ}}\end{array}$$

Two positively charged spheres, each with a charge of \(2.0 \times\) \(10^{-5} \mathrm{C},\) a mass of 1.0 \(\mathrm{kg}\) , and separated by a distance of \(1.0 \mathrm{cm},\) are held in place on a frictionless track. (a) What is the electrostatic potential energy of this system? If the spheres are released, will they move toward or away from each other? (c) What speed will each sphere attain as the distance between the spheres approaches infinity? [Section 5.1\(]\)

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