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Chapter 6: Problem 77

A quantity of \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.862 \mathrm{MHCl}\) is mixed with an equal volume of \(0.431 M \mathrm{Ba}(\mathrm{OH})_{2}\) in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the \(\mathrm{HCl}\) and \(\mathrm{Ba}(\mathrm{OH})_{2}\) solutions is the same at \(20.48^{\circ} \mathrm{C}\), For the process $$\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$$ the heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\). What is the final temperature of the mixed solution?

Short Answer

Expert verified
The final temperature of the mixed solution is 17.58°C.

Step by step solution

01

Identify the limiting reactant

First, you need to find out the number of moles of each reactant. Use the formula \(M = n/V\) where \(M\) is molarity, \(n\) is number of moles, and \(V\) is volume. For HCl: \(n_{HCl} = MV = (0.862 mol/L) \times (2.00 \times 10^{2} ml × 1L/1000ml) = 0.1724 mol\), and for Ba(OH)2: \(n_{Ba(OH)2} = MV = (0.431 mol/L) \times (2.00 \times 10^{2} ml × 1L/1000ml) = 0.0862 mol\). Each HCl molecule reacts with one molecule of Ba(OH)2, so Ba(OH)2 is the limiting reactant.
02

Calculate the heat generated

Next, compute the heat generated using the heat of neutralization and the amount of moles of the limiting reactant. Use the formula \(q = n\Delta H\) where \(q\) is the total heat energy, \(n\) is the number of moles and \(\Delta H\) is the heat of reaction. Since the heat of neutralization is given as -56.2 kJ/mol, a negative value signifies heat is released: \(q = n_{Ba(OH)2}\Delta H = 0.0862 mol \times -56.2 kJ/mol = -4.8436 kJ\).
03

Determine the final temperature

Finally, calculate the final temperature. Totally, 400 mL of solution are obtained by mixing two 200 mL solutions. Assuming the density of the resulting solution is 1 g/mL and its specific heat is 4.18 J/g°C, akin to water, use the formula \(q = mc\Delta T\) where \(m\) is the mass of the solution, \(c\) is its specific heat capacity and \(\Delta T\) is the change in temperature, to find the temperature change: \(\Delta T = q/(mc) = -4.8436 kJ / (400 g \times 4.18 J/g°C)= -2.9°C\). To get the final temperature, add this change to the initial temperature of 20.48°C: \(T_{final} = T_{initial} + \Delta T = 20.48°C - 2.9°C = 17.58°C\).

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Most popular questions from this chapter

The internal energy of an ideal gas depends only on its temperature. Do a first-law analysis of this process. A sample of an ideal gas is allowed to expand at constant temperature against atmospheric pressure. (a) Does the gas do work on its surroundings? (b) Is there heat exchange between the system and the surroundings? If so, in which direction? (c) What is \(\Delta U\) for the gas for this process?

What is heat? How does heat differ from thermal energy? Under what condition is heat transferred from one system to another?

Determine the standard enthalpy of formation of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) from its standard enthalpy of combustion \((-1367.4 \mathrm{~kJ} / \mathrm{mol})\)

Calculate the internal energy of a Goodyear blimp filled with helium gas at \(1.2 \times 10^{5} \mathrm{~Pa}\). The volume of the blimp is \(5.5 \times 10^{3} \mathrm{~m}^{3}\). If all the energy were used to heat 10.0 tons of copper at \(21^{\circ} \mathrm{C},\) calculate the final temperature of the metal. (Hint: See Section 5.7 for help in calculating the internal energy of a gas. 1 ton \(=9.072 \times 10^{5} \mathrm{~g} .\)

A person ate 0.50 pound of cheese (an energy intake of \(4000 \mathrm{~kJ}\) ). Suppose that none of the energy was stored in his body. What mass (in grams) of water would he need to perspire in order to maintain his original temperature? (It takes \(44.0 \mathrm{~kJ}\) to vaporize 1 mole of water.
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