Chapter 6: Problem 2
What is meant by the term lower in energy? Which is lower in energy, a mixture of hydrogen and oxygen gases or liquid water? How do you know? Which of the two is more stable? How do you know?
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Chapter 6: Problem 2
What is meant by the term lower in energy? Which is lower in energy, a mixture of hydrogen and oxygen gases or liquid water? How do you know? Which of the two is more stable? How do you know?
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Consider 5.5 \(\mathrm{L}\) of a gas at a pressure of 3.0 \(\mathrm{atm}\) in a cylinder with a movable piston. The external pressure is changed so that the volume changes to 10.5 \(\mathrm{L}\) . a. Calculate the work done, and indicate the correct sign. b. Use the preceding data but consider the process to occur in two steps. At the end of the first step, the volume is 7.0 \(\mathrm{L}\) . The second step results in a final volume of 10.5 \(\mathrm{L}\) . Calculate the work done, and indicate the correct sign. c. Calculate the work done if after the first step the volume is 8.0 \(\mathrm{L}\) and the second step leads to a volume of 10.5 \(\mathrm{L}\) . Does the work differ from that in part b? Explain.
A gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. Given the following data, determine \(\Delta H_{f}^{\circ}\) for the hydrocarbon: $$ \begin{aligned} \Delta H_{\mathrm{reacion}}^{\circ} &=-2044.5 \mathrm{kJ} / \mathrm{mol} \text { hydrocarbon } \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{CO}_{2}\right) &=-393.5 \mathrm{kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right) &=-242 \mathrm{kJ} / \mathrm{mol} \end{aligned} $$ Density of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) product mixture at 1.00 \(\mathrm{atm}\) , \(200 . \mathrm{C}=0.751 \mathrm{g} / \mathrm{L}\) . The density of the hydrocarbon is less than the density of Kr at the same conditions.
A gas absorbs 45 kJ of heat and does 29 kJ of work. Calculate \(\Delta E .\)
In a coffee-cup calorimeter, 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M}\) NaOH and 100.0 \(\mathrm{mL}\) of 1.0 \(\mathrm{M} \mathrm{HCl}\) are mixed. Both solutions were originally at \(24.6^{\circ} \mathrm{C}\) . After the reaction, the final temperature is \(31.3^{\circ} \mathrm{C}\) . Assuming that all the solutions have a density of 1.0 \(\mathrm{g} / \mathrm{cm}^{3}\) and a specific heat capacity of \(4.18 \mathrm{J} / \mathrm{C} \cdot \mathrm{g},\) calculate the enthalpy change for the neutralization of \(\mathrm{HCl}\) by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter.
The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area \((1 \text { watt }=1 \mathrm{Js} \text { ). The plants in an }\) agricultural field produce the equivalent of \(20 . \mathrm{kg}\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per hectare \(\left(1 \mathrm{ha}=10,000 \mathrm{m}^{2}\right) .\) Assuming that sucrose is produced by the reaction $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+& 12 \mathrm{O}_{2}(g) \\ & \Delta H=5640 \mathrm{kJ} \end{aligned} $$ calculate the percentage of sunlight used to produce the sucrose-that is, determine the efficiency of photosynthesis.
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