Problem 525
Calculate \(\Delta \mathrm{S}\) for the conversion of one mole of liquid water to vapor at \(100^{\circ} \mathrm{C}\). Heat of vaporization \(=540 \mathrm{cal} / \mathrm{g}\).
Problem 527
Determine the entropy change that takes place when 1 mole of ammonia (a) Passes from the liquid state to the gaseous state at its boiling point, \(-33^{\circ} \mathrm{C} ; \Delta \mathrm{H}_{\mathrm{VAP}}=5570 \mathrm{cal} / \mathrm{mole}\) (b) as a gas at \(-33^{\circ} \mathrm{C}\) comes to room temperature, \(25^{\circ} \mathrm{C}\). Assume heat capacity is constant at \(8.9\) cal/deg - mole for this range.
Problem 528
Given, for acetic acid that \(\Delta \mathrm{H}_{\text {fus }}=2592\) cal/mole at its melting point, \(16.6^{\circ} \mathrm{C}\) and \(\Delta \mathrm{H}_{\mathrm{VAP}}=5808 \mathrm{cal} / \mathrm{mole}\) at its boiling point, \(118.3^{\circ} \mathrm{C}\), calculate the change in entropy that takes place when 1 mole of the vapor is condensed at its boiling point and changed to a solid at its melting point, all under constant pressure, taken as 1 atm. Assume that the molar heat capacity of acetic acid is \(27.6 \mathrm{cal} /\) deg \(-\) mole.
Problem 529
Hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}\), can be synthesized in two ways. The first method involves reduction of oxygen by hydrogen, $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}(\ell) $$ The second method involves oxidation of water: $$ 2 \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}_{2}(\ell) $$ Find the free energy of formation, \(\Delta \mathrm{G}^{\circ}\), for both processes and predict which process is more efficient for the commercial preparation of hydrogen peroxide.
Problem 530
Determine \(\Delta \mathrm{G}^{\circ}\) for the reaction \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\ell)\) \(\Delta \mathrm{G}^{\circ} \mathrm{f}\) of \(\mathrm{NH}_{3}(\mathrm{~g})=-4.0 \mathrm{Kcal} / \mathrm{mole}\) \(\Delta \mathrm{G}^{\circ} \mathrm{f}\) of \(\mathrm{NO}(\mathrm{g})=20.7 \mathrm{Kcal} / \mathrm{mole}\) \(\Delta \mathrm{G}^{\circ} \mathrm{f}\) of \(\mathrm{H}_{2} \mathrm{O}(\ell)=-56.7 \mathrm{Kcal} / \mathrm{mole}\)
Problem 531
For sublimation of iodine crystals, $$ \mathrm{I}_{2}(\mathrm{~s}) \rightleftarrows \mathrm{I}_{2}(\mathrm{~g}) $$ at \(25^{\circ} \mathrm{C}\) and atmospheric pressure, it is found that the change in enthalpy, \(\Delta \mathrm{H}=9.41 \mathrm{Kcal} /\) mole and the change in entropy, \(\Delta \mathrm{S}=20.6 \mathrm{cal} /\) deg \(-\) mole. At what temperature will solid iodine be in equilibrium with gaseous iodine?
Problem 532
At the melting point of a solid (or the freezing point of a liquid), the free energies of the solid state and the liquid state are equal, \(\Delta \mathrm{G}=0 .\) Likewise, at the boiling point of a liquid, where there is an equilibrium between the liquid and vapor phases, the free energy is equal in the two states. Calculate the change in entropy for the following process at \(0^{\circ} \mathrm{C}\) if the heat of fusion of \(\mathrm{H}_{2} \mathrm{O}=80 \mathrm{cal} / \mathrm{g} . \mathrm{H}_{2} \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)\).
Problem 534
Determine the entropy difference between solid and liquid states for a substance melting at \(100^{\circ} \mathrm{C}\) and having a heat of fusion of \(10,000 \mathrm{~J} / \mathrm{mol}\)
Problem 538
Determine the free energy change, \(\Delta \mathrm{G}\), for transforming liquid water at \(100^{\circ} \mathrm{C}\) and 1 atm. to vapor at the same conditions. \(\Delta \mathrm{H}=9720\) cal (molar enthalpy of vaporization).
Problem 540
Assuming \(\Delta \mathrm{H}^{\circ}\) remains constant, calculate the equilibrium constant, \(\mathrm{K}\), at \(373^{\circ} \mathrm{K}\), if it equals \(1.6 \times 10^{12}\) at \(298^{\circ} \mathrm{K}\) for the reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{~g})\). The standard enthalpy change for this reaction is \(-113 \mathrm{~kJ} / \mathrm{mole}\).
