P14B.8 Consider the arrangement shown in 5 for a system consisting of an O-H group and an O atom, and then use the electrostatic model of the hydrogen bond to calculate the dependence of the molar potential energy of interaction on the angle 0. -0.83e 200 pm H -0.83e +0.45e 95.7 pm
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- Elaborate four (4) components of spontaneous potential.Calculate the Gibbs free energy (in kJ/mol) at 25 degrees C of the following reaction using the data in the table. 4A +5B → 4C + 2D + 1E ΔHf0(kJ/mol) S0(J/mol/K) A 91.27 181.59 B -86.74 193.27 C 33.09 197.70 D 48.30 55.82 E -78.06 163.42NB Assume the densities of all solutions are 1.0 g/mL and their specific heat capacities 4.184 J/g .K Molar Mass of KCl: 74.55 g/mol Help from calculating qr to average ethalpy.
- What can you conclude given that the plot of ln P versus t islinear?Which of the following conditions is sufficient alone to predict a reaction to be spontaneous in the forward direction? A. Qc > Kc B. IP = Ksp C. ΔG rxn < 0 D. ΔS rxn > 0 E. ΔH rxn < 04. An isolated system contains only 4 particles (N = 4) and total energy E of 4ε. If the only possible energy levels are (0, 2ε, 4ε).a) Give all the possible energy configurationsb) Identify the most probable configurationc) Calculate the entropy of this configuration for the system
- P2C.6 For the reaction Cr(C6H6)2(s) → Cr(s) + 2C6H6(g), ΔrU⦵(583K) = +8.0 kJmol−1. Find the corresponding reaction enthalpy and estimate thestandard enthalpy of formation of Cr(C6H6)2(s) at 583K.Hydrogen is one of only seven elements which exist as stable diatomic molecules at (or close to) room temperature and atmospheric pressure. Let’s investigate just how much more thermodynamically favorable diatomic hydrogen is compared to atomic hydrogen. Given the following reaction and associated data at T = 298.15 K. 2 H(g) ⇌ H"(g) or equivalently H(g) + H(g) ⇌ H2(g) Δf?° (kJ mol-1) ?° (kJ K-1 mol-1) H(g) 218.0 0.115 H2(g) 0 0.131 Calculate ΔH, ΔS, and ΔG for the formation of H2(g) from H(g) at 298.15 K. Calculate KP for the reaction. Calculate the temperature at which the reverse reaction becomes favorable. Assume ΔH and ΔS do not change with temperature.Now consider the London interaction between the phenyl groups of two Phe residues (see Problem P14B.5). (a) Estimate the potential energy of interaction between two such rings (treated as benzene molecules) separated by 0.4 nm. For the ionization energy, use I = 5.0 eV. (b) Given that force is the negative slope of the potential, calculate the distance-dependence of the force acting between two non-bonded groups of atoms, such as the phenyl groups of Phe, in a polypeptide chain that can have a London dispersion interaction with each other. What is the separation at which the force between the phenyl groups (treated as benzene molecules) of two Phe residues is zero? Hint: Calculate the slope by considering the potential energy at r and r + δr, with δr << r, and evaluating {V(r + δr) − V(r)}/δr. At the end of the calculation, let δr become vanishingly small.
- The standard entropies at 298 K for certain group 4A elementsare: C(s, diamond) = 2.43 J/mol-K, Si1s2 = 18.81 J/mol-K, Ge1s2 = 31.09 J/mol-K, and Sn1s2 = 51.818 J/mol-K. All but Sn have the same (diamond) structure. Howdo you account for the trend in the S° values?Do you expect the constant ,1 in (9.30) describing the susceptibility of the covalent bond to be positive or negative? Why?Thermodynamics. Prove the relationship. (dP/dT)G = S/V