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Chapter 6: Problem 65

In a coffee-cup calorimeter, 50.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{AgNO}_{3}\) and 50.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HCl}\) are mixed to yield the following reaction: $$\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s)$$ The two solutions were initially at \(22.60^{\circ} \mathrm{C}\) , and the final temperature is \(23.40^{\circ} \mathrm{C}\) Calculate the heat that accompanies this reacture in kJ/mol of AgCl formed. Assume that the combined solution has a mass of 100.0 \(\mathrm{g}\) and a specific heat capacity of 4.18 \(\mathrm{J} / \rho \mathrm{C} \cdot \mathrm{g} .\)

Short Answer

Expert verified
The heat that accompanies this reaction in the formation of AgCl is 66.88 kJ/mol.

Step by step solution

01

Calculate the moles of reactants

First, we need to find the moles of both Ag⁺ and Cl⁻ in the solutions using the volume and concentration. Moles of Ag⁺: Moles = Molarity × Volume Moles of Ag⁺ = 0.100 M × 50.0 mL = 0.100 mol/L × 0.050 L = 0.005 mol Moles of Cl⁻: Moles of Cl⁻ = 0.100 M × 50.0 mL = 0.100 mol/L × 0.050 L = 0.005 mol
02

Determine the limiting reactant

In this case, there are equal moles of Ag⁺ and Cl⁻, and the stoichiometry of the reaction is 1:1. Therefore, both reactants are limiting, and there will be 0.005 mol of AgCl formed.
03

Calculate the heat released

The heat (q) released during the reaction can be calculated using the formula: q = mcΔT where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. In our case, the mass of the combined solution is 100.0 g, the specific heat capacity is 4.18 J/g°C, and the change in temperature is the final temperature minus the initial temperature (ΔT = 23.40°C - 22.60°C). q = (100.0 g)(4.18 J/g°C)(23.40 °C - 22.60 °C) q = (100.0 g)(4.18 J/g°C)(0.80 °C) q = 334.4 J
04

Calculate the heat released per mole of AgCl formed

Now we need to calculate the heat released in kJ/mol of AgCl. We have found that 0.005 moles of AgCl are formed, and the heat released is 334.4 J. We can convert the heat into kJ and then divide by the moles of AgCl formed. q = 334.4 J × (1 kJ / 1000 J) = 0.3344 kJ Heat released per mole of AgCl formed = Heat released / Moles of AgCl formed Heat released per mole of AgCl formed = 0.3344 kJ / 0.005 mol = 66.88 kJ/mol Solution: The heat that accompanies this reaction in the formation of AgCl is 66.88 kJ/mol.

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Most popular questions from this chapter

Given the following data $$ \begin{array}{ll}{\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-23 \mathrm{kJ}} \\ {3 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=-39 \mathrm{kJ}} \\ {\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow 3 \mathrm{FeO}(s)+\mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=18 \mathrm{kJ}}\end{array} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{FeO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$

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In a bomb calorimeter, the reaction vessel is surrounded by water that must be added for each experiment. Since the amount of water is not constant from experiment to experiment, the mass of water must be measured in each case. The heat capacity of the calorimeter is broken down into two parts: the water and the calorimeter components. If a calorimeter contains 1.00 \(\mathrm{kg}\) water and has a total heat capacity of \(10.84 \mathrm{kJ} / \mathrm{C},\) what is the heat capacity of the calorimeter components?

A system absorbs 35 \(\mathrm{J}\) of heat and has 25 \(\mathrm{J}\) of work performed on it. The system then returns to its initial state by a second step. If 5 \(\mathrm{J}\) of heat are given off in the second step, how much work is done by the system in the second step?

On Easter Sunday, April \(3,1983,\) nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: $$ 2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g) $$ a. Calculate \(\Delta H^{\circ}\) for this reaction. Approximately \(2.0 \times\) \(10^{4}\) gal nitric acid was spilled. Assume that the acid was an aqueous solution containing 70.0\(\% \mathrm{HNO}_{3}\) by mass with a density of 1.42 \(\mathrm{g} / \mathrm{cm}^{3} .\) What mass of sodium car- bonate was required for complete neutralization of the spill, and what quantity of heat was evolved? ( \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{NaNO}_{3}(a q)=-467 \mathrm{kJ} / \mathrm{mol} )\) b. According to The Denver Post for April \(4,1983,\) authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of \(\Delta H^{\circ},\) what was their major concern?
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