Chapter 5: Problem 74
Why is it important for Hess's law that enthalpy is a state function?
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Chapter 5: Problem 74
Why is it important for Hess's law that enthalpy is a state function?
These are the key concepts you need to understand to accurately answer the question.
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When measuring the enthalpy of combustion of a very small amount of material, would you prefer to use a calorimeter having a heat capacity that is small or large? Explain your reasoning.
Acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\left(\Delta H_{\mathrm{f}}^{\circ}=226.7 \mathrm{kJ} / \mathrm{mol}\right),\) and benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}\left(\Delta H_{\mathrm{f}}^{\circ}=49.0 \mathrm{kJ} / \mathrm{mol}\right),\) are sometimes referred to as endothermic compounds. a. Why are \(\mathrm{C}_{2} \mathrm{H}_{2}\) and \(\mathrm{C}_{6} \mathrm{H}_{6}\) called endothermic compounds? b. Calculate the standard molar enthalpy of combustion of acetylene and benzene.
Burning \(1.43 \mathrm{g}\) of methane \(\left[\mathrm{CH}_{4}(g), \text { molar mass } 16.0 \mathrm{g} /\right.\) mol] with excess \(\mathrm{O}_{2}\) raised the temperature of \(250 \mathrm{g}\) of water in a bomb calorimeter from \(22^{\circ} \mathrm{C}\) to \(98^{\circ} \mathrm{C}\). Calculate the molar enthalpy of combustion of methane.
Which of the following substances has a standard enthalpy of formation equal to zero? (a) Pb at \(1000^{\circ} \mathrm{C} ;\) (b) \(\mathrm{C}_{3} \mathrm{H}_{8}(g)\) at \(25.0^{\circ} \mathrm{C}\) and 1 atm pressure; (c) solid glucose at room temperature; (d) \(\mathrm{N}_{2}(g)\) at \(25.0^{\circ} \mathrm{C}\) and 1 atm pressure.
The reactor-core cooling systems in some nuclear power plants use liquid sodium as the coolant. Sodium has a thermal conductivity of \(1.42 \mathrm{J} /\) \((\mathrm{cm} \cdot \mathrm{s} \cdot \mathrm{K}),\) which is quite high compared with that of water \(\left.\left[6.1 \times 10^{-3} \mathrm{J} / \mathrm{cm} \cdot \mathrm{s} \cdot \mathrm{K}\right)\right] .\) The respective molar heat capacities are \(28.28 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K})\) and \(75.31 \mathrm{J} /(\mathrm{mol} \cdot \mathrm{K}) .\) What is the advantage of using liquid sodium over water in this application?
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