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Chapter 7: Problem 26

The heat of neutralization of \(\mathrm{HCl}(\text { aq) by } \mathrm{NaOH}(\mathrm{aq})\) is \(-55.84 \mathrm{kJ} / \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) produced. If \(50.00 \mathrm{mL}\) of \(1.05 \mathrm{M}\) \(\mathrm{NaOH}\) is added to \(25.00 \mathrm{mL}\) of \(1.86 \mathrm{M} \mathrm{HCl}\), with both solutions originally at \(24.72^{\circ} \mathrm{C},\) what will be the final solution temperature? (Assume that no heat is lost to the surrounding air and that the solution produced in the neutralization reaction has a density of \(1.02 \mathrm{g} / \mathrm{mL}\) and a specific heat of \(3.98 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

Short Answer

Expert verified
The final temperature of the solution will depend on the given values for volume, molarity, and specific heat, as well as the heat of neutralization. Use the approach outlined in the solution steps to calculate it.

Step by step solution

01

Determine the limiting reactant

In this reaction, hydrochloric acid (\(HCl\)) and sodium hydroxide (\(NaOH\)) react to form water and salt. The amount of heat produced depends on the amount of water formed, which in turn depends on the limiting reactant. Draw up a table with the volume and molarity of each reactant to calculate the number of moles of each reactant. The reactant with fewer moles is the limiting reactant.
02

Calculate the heat released by the reaction

Using the heat of neutralization as given (-55.84 kJ/mol of \(H_{2}O\) produced), and the moles of the limiting reactant (the one that generates water), calculate the total heat released by the reaction. Remember the output will need to be converted to joules (1 kJ = 1000 J) to match the required units in subsequent steps.
03

Determine the mass of the solution

Next, calculate the total mass of the final solution using the given densities and the sum of the initial volumes of the reactants. Remember that density = mass/volume and rearrange the formula to find the mass of the solution.
04

Calculate the final temperature

Now you can find the final temperature of the solution. The amount of heat absorbed by the solution will equal the amount of heat released by the reaction. This heat can be calculated using the equation \(q=mc\Delta T\), where m is the mass of the solution, c is its specific heat, and \(\Delta T\) is the temperature difference. Rearrange this formula to find \(\Delta T = q / (mc)\), then add this to the initial temperature to find the final temperature of the solution.

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Most popular questions from this chapter

Upon complete combustion, the indicated substances evolve the given quantities of heat. Write a balanced equation for the combustion of \(1.00 \mathrm{mol}\) of each substance, including the enthalpy change, \(\Delta H\), for the reaction.Upon complete combustion, the indicated substances evolve the given quantities of heat. Write a balanced equation for the combustion of \(1.00 \mathrm{mol}\) of each substance, including the enthalpy change, \(\Delta H\), for the reaction. (a) \(0.584 \mathrm{g}\) of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) yields \(29.4 \mathrm{kJ}\) (b) \(0.136 \mathrm{g}\) of camphor, \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}(\mathrm{s}),\) yields \(5.27 \mathrm{kJ}\) (c) \(2.35 \mathrm{mL}\) of acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(\mathrm{l})(d=0.791\) \(\mathrm{g} / \mathrm{mL}),\) yields \(58.3 \mathrm{kJ}\)

A \(1.00 \mathrm{g}\) sample of \(\mathrm{Ne}(\mathrm{g})\) at 1 atm pressure and \(27^{\circ} \mathrm{C}\) is allowed to expand into an evacuated vessel of \(2.50 \mathrm{L}\) volume. Does the gas do work? Explain.

Briefly describe each of the following ideas or methods: (a) law of conservation of energy; (b) bomb calorimetry; (c) function of state; (d) enthalpy diagram; (e) Hess's law.

Hot water and a piece of cold metal come into contact in an isolated container. When the final temperature of the metal and water are identical, is the total energy change in this process (a) zero; (b) negative; (c) positive; (d) not enough information.

A 1.620 g sample of naphthalene, \(C_{10} \mathrm{H}_{8}(\mathrm{s}),\) is completely burned in a bomb calorimeter assembly and a temperature increase of \(8.44^{\circ} \mathrm{C}\) is noted. If the heat of combustion of naphthalene is \(-5156 \mathrm{kJ} / \mathrm{mol} \mathrm{C}_{10} \mathrm{H}_{8}\) what is the heat capacity of the bomb calorimeter?
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