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Chapter 6: Problem 115

An excess of zinc metal is added to \(50.0 \mathrm{~mL}\) of a \(0.100 M\) AgNO \(_{3}\) solution in a constant-pressure calorimeter like the one pictured in Figure \(6.9 .\) As a result of the reaction $$\mathrm{Zn}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 \mathrm{Ag}(s)$$ the temperature rises from \(19.25^{\circ} \mathrm{C}\) to \(22.17^{\circ} \mathrm{C}\). If the heat capacity of the calorimeter is \(98.6 \mathrm{~J} /{ }^{\circ} \mathrm{C},\) calculate the enthalpy change for the above reaction on a molar basis. Assume that the density and specific heat of the solution are the same as those for water, and ignore the specific heats of the metals.

Short Answer

Expert verified
The enthalpy change for the reaction is calculated by calculating the total heat absorbed by the solution and calorimeter and then converting this heat into molar enthalpy using the stoichiometry of the reaction and the initial concentration of the reactant (AgNO3).

Step by step solution

01

Calculate Heat Absorbed by Solution

First, calculate the heat absorbed by the solution using the equation: \[ q = m \cdot c \cdot \Delta T\] where \(m\) is the mass of the solution (assume as equal to the mass of water, as density of solution and water are said to be equal), \(c\) is the specific heat capacity of water (4.184 J/g°C), and \(\Delta T\) is the change in temperature. Given that the solution volume is 50.0mL, which equals 50.0g for water, and the temperature change is 22.17°C - 19.25°C, calculate the heat absorbed by the solution.
02

Calculate Heat Absorbed by Calorimeter

Next, calculate the heat absorbed by the calorimeter using the equation: \[ q_{\text{cal}} = C_{\text{cal}} \cdot \Delta T\] where \(C_{\text{cal}}\) is the heat capacity of the calorimeter (98.6 J/°C) and \(\Delta T\) is the change in temperature. Then calculate the total heat absorbed as: \[ q_{\text{total}} = q_{\text{sol}} + q_{\text{cal}}\]
03

Convert Heat into Molar Enthalpy

For converting the total heat absorbed into molar enthalpy , divide the total heat absorbed (negative as it is absorbed) by the number of moles of Zn reacted (which can be calculated based on the original 0.100 M concentration of AgNO3 and the reaction stoichiometry).

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Most popular questions from this chapter

Consider the following data: $$ \begin{array}{lcc} \text { Metal } & \text { Al } & \text { Cu } \\ \hline \text { Mass (g) } & 10 & 30 \\ \text { Specific heat (J/g* } \left.^{\circ} \mathrm{C}\right) & 0.900 & 0.385 \\\ \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & 40 & 60 \\ \hline \end{array} $$ When these two metals are placed in contact, which of the following will take place? (a) Heat will flow from Al to Cu because Al has a larger specific heat. (b) Heat will flow from \(\mathrm{Cu}\) to \(\mathrm{Al}\) because \(\mathrm{Cu}\) has a larger mass. (c) Heat will flow from Cu to Al because Cu has a larger heat capacity. (d) Heat will flow from Cu to Al because Cu is at a higher temperature. (e) No heat will flow in either direction.

Ice at \(0^{\circ} \mathrm{C}\) is placed in a Styrofoam cup containing \(361 \mathrm{~g}\) of a soft drink at \(23^{\circ} \mathrm{C}\). The specific heat of the drink is about the same as that of water. Some ice remains after the ice and soft drink reach an equilibrium temperature of \(0^{\circ} \mathrm{C}\). Determine the mass of ice that has melted. Ignore the heat capacity of the cup.

A 44.0-g sample of an unknown metal at \(99.0^{\circ} \mathrm{C}\) was placed in a constant-pressure calorimeter containing \(80.0 \mathrm{~g}\) of water at \(24.0^{\circ} \mathrm{C}\). The final temperature of the system was found to be \(28.4^{\circ} \mathrm{C}\). Calculate the specific heat of the metal. (The heat capacity of the calorimeter is \(\left.12.4 \mathrm{~J} /{ }^{\circ} \mathrm{C} .\right)\)

A 46-kg person drinks 500 g of milk, which has a "caloric" value of approximately \(3.0 \mathrm{~kJ} / \mathrm{g}\). If only 17 percent of the energy in milk is converted to mechanical work, how high (in meters) can the person climb based on this energy intake? [Hint: The work done in ascending is given by \(m g h,\) where \(m\) is the mass (in kilograms), \(g\) the gravitational acceleration \(\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right),\) and \(h\) the height (in meters).]

Metabolic activity in the human body releases approximately \(1.0 \times 10^{4} \mathrm{~kJ}\) of heat per day. Assuming the body is \(50 \mathrm{~kg}\) of water, how much would the body temperature rise if it were an isolated system? How much water must the body eliminate as perspiration to maintain the normal body temperature \(\left(98.6^{\circ} \mathrm{F}\right) ?\) Comment on your results. The heat of vaporization of water may be taken as \(2.41 \mathrm{~kJ} / \mathrm{g}\).
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