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Chapter 5: Problem 45

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

Short Answer

Expert verified
(a) The enthalpy change for the production of 0.450 moles of silver chloride is -29.475 kJ. (b) The enthalpy change for the production of 9.00 g of silver chloride is approximately -4.113 kJ. (c) The enthalpy change for the dissolution of 9.25 x 10^-4 mol of silver chloride is approximately 0.0606 kJ.

Step by step solution

01

(a) Enthalpy change for 0.450 mol AgCl production

To find the enthalpy change for this specific situation, we can use the direct proportionality between enthalpy change and moles of silver chloride produced: \(\Delta H = -65.5\ \mathrm{kJ/mol} \times 0.450\ \mathrm{mol}\) \(\Delta H = -29.475\ \mathrm{kJ}\) The enthalpy change for the production of 0.450 moles of silver chloride is -29.475 kJ.
02

(b) Enthalpy change for 9.00 g AgCl production

Firstly, let's convert the mass of AgCl to moles, given that the molar mass of AgCl = 143.32 g/mol. Moles of AgCl: \(n = \frac{9.00\ \mathrm{g}}{143.32\ \mathrm{g/mol}}\) n ≈ 0.0628 mol Now, we can calculate the enthalpy change for the production of 9.00 g (0.0628 mol) of silver chloride: \(\Delta H = -65.5\ \mathrm{kJ/mol} \times 0.0628\ \mathrm{mol}\) \(\Delta H ≈ -4.113\ \mathrm{kJ}\) The enthalpy change for the production of 9.00 g of silver chloride is approximately -4.113 kJ.
03

(c) Enthalpy change for 9.25 x 10^-4 mol AgCl dissolution

As the enthalpy change given corresponds to the formation of solid AgCl from aqueous Ag+ and Cl-, we need to reverse the process for dissolution of AgCl back into its soluble ions. So, the sign of enthalpy change will be reversed. Now, we can find the enthalpy change for the dissolution of 9.25 x 10^-4 mol of AgCl: \(\Delta H = 65.5\ \mathrm{kJ/mol} \times 9.25 \times 10^{-4}\ \mathrm{mol}\) \(\Delta H ≈ 0.0606\ \mathrm{kJ}\) The enthalpy change for the dissolution of 9.25 x 10^-4 mol of silver chloride is approximately 0.0606 kJ.

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Most popular questions from this chapter

The corrosion (rusting) of iron in oxygen-free water includes the formation of iron(II) hydroxide from iron by the following reaction: $$ \mathrm{Fe}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2}(g) $$ If 1 mol of iron reacts at \(298 \mathrm{~K}\) under \(101.3 \mathrm{kPa}\) pressure, the reaction performs \(2.48 \mathrm{~J}\) of \(P-V\) work, pushing back the atmosphere as the gaseous \(\mathrm{H}_{2}\) forms. At the same time, $11.73 \mathrm{~kJ}$ of heat is released to the environment. What are the values of \(\Delta H\) and of \(\Delta E\) for this reaction?

Butane \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) boils at $-0.5^{\circ} \mathrm{C} ;\( at this temperature it has a density of \)0.60 \mathrm{~g} / \mathrm{cm}^{3}\(. The enthalpy of formation of \)\mathrm{C}_{4} \mathrm{H}_{10}(g)\( is \)-124.7 \mathrm{~kJ} / \mathrm{mol},$ and the enthalpy of vaporiza- tion of \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) is $22.44 \mathrm{~kJ} / \mathrm{mol} .\( Calculate the enthalpy change when \)1 \mathrm{~L}$ of liquid \(\mathrm{C}_{4} \mathrm{H}_{10}(l)\) is burned in air to give \(\mathrm{CO}_{2}(g)\) 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}(l),\( the density at \)25^{\circ} \mathrm{C}\( is \)0.792 \mathrm{~g} / \mathrm{cm}^{3},\( and \)\Delta H_{f}^{\circ}=-239 \mathrm{~kJ} / \mathrm{mol}$.

(a) What is the electrostatic potential energy (in joules) between an electron and a proton that are separated by \(230 \mathrm{pm}\) ? (b) What is the change in potential energy if the distance separating the electron and proton is increased to \(1.0 \mathrm{nm}\) ? (c) Does the potential energy of the two particles increase or decrease when the distance is increased to \(1.0 \mathrm{nm}\) ?

Without referring to tables, predict which of the following has the higher enthalpy in each case: (a) \(1 \mathrm{~mol} \mathrm{I}_{2}(s)\) or $1 \mathrm{~mol} \mathrm{I}_{2}(g)$ at the same temperature, (b) \(2 \mathrm{~mol}\) of iodine atoms or \(1 \mathrm{~mol}\) of \(\mathrm{I}_{2},(\mathbf{c}) 1 \mathrm{~mol} \mathrm{I}_{2}(g)\) and $1 \mathrm{~mol} \mathrm{H}_{2}(g)$ at \(25^{\circ} \mathrm{C}\) or \(2 \mathrm{~mol} \mathrm{HI}(g)\) at $25^{\circ} \mathrm{C},(\mathbf{d}) 1 \mathrm{~mol} \mathrm{H}_{2}(g)\( at \)100^{\circ} \mathrm{C}$ or \(1 \mathrm{~mol} \mathrm{H}_{2}(g)\) at \(300^{\circ} \mathrm{C}\).

The hydrocarbons cyclohexane $\left(\mathrm{C}_{6} \mathrm{H}_{12}(l), \Delta H_{f}^{\circ}=-156\right.$ \(\mathrm{kJ} / \mathrm{mol}\) ) and 1-hexene $\left(\mathrm{C}_{6} \mathrm{H}_{12}(l), \Delta H_{f}^{\circ}=-74 \mathrm{~kJ} / \mathrm{mol}\right)$ have the same empirical formula. (a) Calculate the standard enthalpy change for the transformation of cyclohexane to 1-hexene. (b) Which has greater enthalpy, cyclohexane or 1-hexene? (c) Without doing a further calculation and knowing the answer to (b), do you expect cyclohexane or 1-hexene to have the larger combustion enthalpy?
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