Chapter 6: Problem 9
Hess's law is really just another statement of the first law of thermodynamics. Explain.
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Chapter 6: Problem 9
Hess's law is really just another statement of the first law of thermodynamics. Explain.
These are the key concepts you need to understand to accurately answer the question.
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In a coffee-cup calorimeter, \(1.60 \mathrm{~g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) is mixed with \(75.0 \mathrm{~g}\) water at an initial temperature of \(25.00^{\circ} \mathrm{C}\). After dissolution of the salt, the final temperature of the calorimeter contents is \(23.34^{\circ} \mathrm{C}\). Assuming the solution has a heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) and assuming no heat loss to the calorimeter, calculate the enthalpy change for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in units of \(\mathrm{kJ} / \mathrm{mol}\).
A gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. Given the following data, determine \(\Delta H_{\mathrm{f}}^{\circ}\) for the hydrocarbon:. $$ \begin{aligned} \Delta H_{\text {reaction }}^{\circ} &=-2044.5 \mathrm{~kJ} / \mathrm{mol} \text { hydrocarbon } \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{CO}_{2}\right) &=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right) &=-242 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ Density of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) product mixture at \(1.00 \mathrm{~atm}\), \(200 .{ }^{\circ} \mathrm{C}=0.751 \mathrm{~g} / \mathrm{L}\) The density of the hydrocarbon is less than the density of \(\mathrm{Kr}\) at the same conditions.
In a coffee-cup calorimeter, \(150.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) is added to \(50.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{NaOH}\) to make \(200.0 \mathrm{~g}\) solution at an initial temperature of \(48.2^{\circ} \mathrm{C}\). If the enthalpy of neutralization for the reaction between a strong acid and a strong base is \(-56 \mathrm{~kJ} / \mathrm{mol}\), calculate the final temperature of the calorimeter contents. Assume the specific heat capacity of the solution is \(4.184 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and assume no heat loss to the surroundings.
Consider the reaction \(2 \mathrm{HCl}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{BaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H=-118 \mathrm{~kJ}\) Calculate the heat when \(100.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{HCl}\) is mixed with \(300.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\). Assuming that the temperature of both solutions was initially \(25.0^{\circ} \mathrm{C}\) and that the final mixture has a mass of \(400.0 \mathrm{~g}\) and a specific heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\), calculate the final temperature of the mixture.
A \(110 .-\mathrm{g}\) sample of copper (specific heat capacity \(=0.20 \mathrm{~J} /{ }^{\circ} \mathrm{C}\). \(\mathrm{g}\) ) is heated to \(82.4^{\circ} \mathrm{C}\) and then placed in a container of water at \(22.3^{\circ} \mathrm{C}\). The final temperature of the water and copper is \(24.9^{\circ} \mathrm{C}\). What is the mass of the water in the container, assum- ing that all the heat lost by the copper is gained by the water?
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