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Chapter 5: Problem 45

When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s) \quad \Delta H=-65.5 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) for 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 AgCl. (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 mol of AgCl is -29.475 kJ. (b) The enthalpy change for the production of 9.00 g of AgCl is -4.119 kJ. (c) The enthalpy change for the dissolution of 9.25 * 10^{-4} mol of AgCl in water is 0.0605375 kJ.

Step by step solution

01

(a) Given moles of AgCl and ΔH of reaction, calculate ΔH for the production of 0.450 mol of AgCl

First, we must determine the molar enthalpy change, as the given ΔH is for the formation of one mole of AgCl. Molar enthalpy change: ΔH = -65.5 kJ/mol Now, we can calculate the enthalpy change for the production of 0.450 mol of AgCl using the molar enthalpy change: ΔH = (molar enthalpy change) × (moles of AgCl) ΔH = -65.5 kJ/mol × 0.450 mol
02

(a) Calculate the enthalpy change

ΔH = -29.475 kJ So the enthalpy change for the production of 0.450 mol of AgCl is -29.475 kJ.
03

(b) Calculate the moles of AgCl, given 9.00 g of AgCl

To find the enthalpy change for the production of 9.00 g of AgCl, we first need to determine the moles of AgCl, using its molar mass: Molar mass of AgCl = 107.87 (Ag) + 35.45 (Cl) = 143.32 g/mol Moles of AgCl = mass / molar mass Moles of AgCl = 9.00 g / 143.32 g/mol
04

(b) Calculate the enthalpy change for the 9.00 g of AgCl

Now, we can calculate the enthalpy change for the production of 9.00 g of AgCl using the molar enthalpy change: ΔH = (molar enthalpy change) × (moles of AgCl) ΔH = -65.5 kJ/mol × (9.00 g / 143.32 g/mol)
05

(b) Calculate the enthalpy change

ΔH = -4.119 kJ So the enthalpy change for the production of 9.00 g of AgCl is -4.119 kJ.
06

(c) Calculate ΔH for the dissolution of 9.25 * 10^{-4} mol of AgCl

For the dissolution of AgCl in water, we need to reverse the given reaction, and thus, the enthalpy change will have the opposite sign: Molar enthalpy change: ΔH = 65.5 kJ/mol Now, we can calculate the enthalpy change for the dissolution of 9.25 * 10^{-4} mol of AgCl in water using the molar enthalpy change: ΔH = (molar enthalpy change) × (moles of AgCl) ΔH = 65.5 kJ/mol × 9.25 * 10^{-4} mol
07

(c) Calculate the enthalpy change

ΔH = 0.0605375 kJ So the enthalpy change for the dissolution of 9.25 * 10^{-4} mol of AgCl in water is 0.0605375 kJ.

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