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Chapter 7: Problem 21

A 0.205 g pellet of potassium hydroxide, \(\mathrm{KOH}\), is added to \(55.9 \mathrm{g}\) water in a Styrofoam coffee cup. The water temperature rises from 23.5 to \(24.4^{\circ} \mathrm{C}\). [Assume that the specific heat of dilute \(\mathrm{KOH}(aq)\) is the same as that of water.] (a) What is the approximate heat of solution of \(\mathrm{KOH}\) expressed as kilojoules per mole of \(\mathrm{KOH}?\) (b) How could the precision of this measurement be improved without modifying the apparatus?

Short Answer

Expert verified
a) The approximate heat of solution of KOH is 57.5 kJ/mol. b) Precision can be improved by ensuring the initial temperature of the KOH pellet and the water is the same, stirring more thoroughly, and doing multiple trials for average value.

Step by step solution

01

Calculating the heat absorbed by the water

Let's calculate the amount of heat absorbed by the water using the formula \(q = mc\Delta T\).\nGiven: mass of water (m) = 55.9 g, Specific heat of water (c) = 4.18 J/g°C, Change in temperature (\(\Delta T\)) = \(24.4 °C - 23.5 °C = 0.9 °C\)\n Substituting these values into the formula gives: \(q = 55.9 \, g \times 4.18 \, J/g°C \times 0.9 °C = 209.8 J\). This is the heat absorbed by the water.
02

Calculate moles of KOH used

The heat absorbed by the water equals the heat of solution of the KOH pellet. To determine the heat of solution per mole of KOH, calculate the number of moles in the 0.205 g pellet. This is done by dividing the mass of the pellet by the molar mass of KOH:\n Molar mass of KOH = 39.1g (for K) + 16.0g (for O) + 1.01g (for H) = 56.11g/mol\n Therefore, the number of moles = 0.205 g / 56.11 g/mol = 0.00365 mol
03

Calculating the heat of solution of KOH

To find the heat of solution of KOH, one must divide the quantity of heat absorbed by the number of moles of KOH:\n Heat of solution (q) = 209.8 J / 0.00365 mol = 57449.32 J/mol, which can be approximated to 57.5 kJ/mol after converting from Joules to kiloJoules.
04

Improving the precision of the measurement

Some ways to improve the precision of the measurement without modifying the apparatus include ensuring that the initial temperature of the KOH pellet and the water is the same to minimize heat exchange prior to mixing, stirring the solution more thoroughly to ensure that the KOH fully dissolves and distributes the heat evenly, and performing the experiment multiple times to obtain an average value.

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Most popular questions from this chapter

The following substances undergo complete combustion in a bomb calorimeter. The calorimeter assembly has a heat capacity of \(5.136 \mathrm{kJ} /^{\circ} \mathrm{C} .\) In each case, what is the final temperature if the initial water temperature is \(22.43^{\circ} \mathrm{C} ?\) \(\begin{array}{lllll}\text { (a) } 0.3268 & \text { g caffeine, } & \mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{N}_{4} & \text { (heat of }\end{array}\) combustion \(=-1014.2 \mathrm{kcal} / \mathrm{mol} \text { caffeine })\) (b) \(1.35 \mathrm{mL}\) of methyl ethyl ketone, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}(1)\) \(d=0.805 \mathrm{g} / \mathrm{mL}\) (heat of combustion \(=-2444 \mathrm{kJ} / \mathrm{mol}\) methyl ethyl ketone).

How much heat, in kilojoules, is evolved in the complete combustion of (a) \(1.325 \mathrm{g} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{atm} ;\) (b) \(28.4 \mathrm{L} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(\mathrm{STP} ;(\mathrm{c})\) \(12.6 \mathrm{LC}_{4} \mathrm{H}_{10}(\mathrm{g})\) at \(23.6^{\circ} \mathrm{C}\) and \(738 \mathrm{mmHg} ?\) Assume that the enthalpy change for the reaction does not change significantly with temperature or pressure. The complete combustion of butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}),\) is represented by the equation $$\begin{array}{r} \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g})+\frac{13}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 4 \mathrm{CO}_{2}(\mathrm{g})+5 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ}=-2877 \mathrm{kJ} \end{array}$$

A \(7.26 \mathrm{kg}\) shot (as used in the sporting event, the shot put) is dropped from the top of a building \(168 \mathrm{m}\) high. What is the maximum temperature increase that could occur in the shot? Assume a specific heat of \(0.47 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\) for the shot. Why would the actual measured temperature increase likely be less than the calculated value?

Given the following information: $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{NH}_{3}(\mathrm{g})\quad\quad\quad\quad\Delta H_{1}^{\circ}$$ $$\mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{NO}(\mathrm{g})+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H_{2}^{\circ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\quad\quad\quad\Delta H_{3}^{\circ}$$ Determine \(\Delta H^{\circ}\) for the following reaction, expressed in terms of \(\Delta H_{1}^{\circ}, \Delta H_{2}^{\circ},\) and \(\Delta H_{3}^{\circ}\) $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}(\mathrm{g}) \quad \Delta H^{\circ}=?$$

The metabolism of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) yields \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) as products. Heat released in the process is converted to useful work with about \(70 \%\) efficiency. Calculate the mass of glucose metabolized by a \(58.0 \mathrm{kg}\) person in climbing a mountain with an elevation gain of \(1450 \mathrm{m}\). Assume that the work performed in the climb is about four times that required to simply lift \(58.0 \mathrm{kg}\) by \(1450 \mathrm{m} \cdot\left(\Delta H_{\mathrm{f}}^{2} \text { of } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}) \text { is }-1273.3 \mathrm{kJ} / \mathrm{mol} .\right)\)
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