Problem 84
Consider the reaction: \(2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)\) When 2 moles of Na react with water at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), the volume of \(\mathrm{H}_{2}\) formed is \(24.5 \mathrm{~L}\). Calculate the work done in joules when \(0.34 \mathrm{~g}\) of Na reacts with water under the same conditions. (The conversion factor is \(1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J} .)\)
Problem 85
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 \(12.4 \mathrm{~J} /{ }^{\circ} \mathrm{C} .\)
Problem 86
A student mixes \(88.6 \mathrm{~g}\) of water at \(74.3^{\circ} \mathrm{C}\) with \(57.9 \mathrm{~g}\) of water at \(24.8^{\circ} \mathrm{C}\) in an insulated flask. What is the final temperature of the combined water?
Problem 87
You are given the following data: \(\begin{aligned} \mathrm{H}_{2}(g) & \longrightarrow 2 \mathrm{H}(g) & & \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Br}_{2}(g) & \longrightarrow 2 \mathrm{Br}(g) & & \Delta H^{\circ}=192.5 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) & \longrightarrow 2 \mathrm{HBr}(g) & \Delta H^{\circ} &=-72.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\) Calculate \(\Delta H^{\circ}\) for the reaction\(\mathrm{H}(g)+\mathrm{Br}(g) \longrightarrow \mathrm{HBr}(g)\).
Problem 89
Ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and gasoline (assumed to be all octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\) ) are both used as automobile fuel. If gasoline is selling for \(\$ 2.20 / \mathrm{gal},\) what would the price of ethanol have to be in order to provide the same amount of heat per dollar? The density and \(\Delta H_{\mathrm{f}}^{\circ}\) of octane are \(0.7025 \mathrm{~g} / \mathrm{mL}\) and \(-249.9 \mathrm{~kJ} / \mathrm{mol}\), respectively, and of ethanol are \(0.7894 \mathrm{~g} / \mathrm{mL}\) and \(-277.0 \mathrm{~kJ} / \mathrm{mol}\) respectively \((1 \mathrm{gal}=3.785 \mathrm{~L})\).
Problem 92
Explain the cooling effect experienced when ethanol is rubbed on your skin, given that \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \quad \Delta H^{\circ}=42.2 \mathrm{~kJ} / \mathrm{mol}\)
Problem 93
For which of the following reactions does \(\Delta H_{\mathrm{rxn}}^{\circ}=\Delta H_{\mathrm{f}}^{\circ}\) ? (a) \(\mathrm{H}_{2}(g)+\mathrm{S}(\) rhombic \() \longrightarrow \mathrm{H}_{2} \mathrm{~S}(g)\) (b) \(\mathrm{C}(\) diamond \()+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)\) (c) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CuO}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cu}(s)\) (d) \(\mathrm{O}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{3}(g)\)
Problem 94
Calculate the work done (in joules) when \(1.0 \mathrm{~mole}\) of water is frozen at \(0^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\). The volumes of 1 mole of water and ice at \(0^{\circ} \mathrm{C}\) are 0.0180 and \(0.0196 \mathrm{~L},\) respectively. (The conversion factor is \(1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J} .)\)
Problem 95
A certain gas initially at \(0.050 \mathrm{~L}\) undergoes expansion until its volume is \(0.50 \mathrm{~L}\). Calculate the work done (in joules) by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 0.20 atm. (The conversion factor is \(1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J} .\) )
Problem 96
Calculate the standard enthalpy of formation for diamond, given that $$ \begin{aligned} \text { C(graphite) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta H^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{C}(\text { diamond })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \\ \Delta H^{\circ} &=-395.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$