3C.3 An FCIO3 molecule can adopt four orientations in the solid with negligible difference in energy. What is its residual molar entropy?

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3) The enzyme aldolase catalyzes the conversion of fructose-1,6-diphosphate (FDP)to dihydroxyacetone phosphate (DHAP) and glyceraldehide-3-phosphate (G3P). Thereaction isFDP ⇄ DHAP + G3P , ∆rG0(298.15 K) = 23.8 kJ mol−1In red blood cells the concentration of these species are [FDP] = 35 µM, [DHAP] =130 µM, and [G3P] = 15 µM. (Remember that 1.0 µM = 1.0 × 10−6 mol L−1. Thestandard state for reactions in solution can be taken as c0 = 1.0 mol L−1).Calculate ∆rG in a red blood cell at 25 0C. Will the reaction occur spontaneouslyin the cell at this temperature?

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.

3. A particular protein heat denatures with a melting temperature Tm = 332 K with an enthalpy of unfolding, AHm = 372 kJmol-¹. (a) What is the entropy of unfolding in kJ mol-1 K-1 at Tm? (b) If ACp is 6.6 kJ mol-1 K-1, at what temperature does the free energy of unfolding (AG unfolding) have the greatest value (i.e. the native state is most stable)? (c) Calculate the value of AG and K at this temperature.

Consider a molecule having three energy levels as Part A follows: What is the probability that this molecule will be in the lowest-energy state? State Energy (cm-1) Degeneracy Express your answer to three significant figures. 1 1 500. 3 ΑΣφ 3 1500. 5 Imagine a collection of N molecules all at 400. K in which one of these molecules is selected. Pi = Note: k = 0.69503476 cm¬1 . K-1. Submit Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining

The second law of thermodynamics states that for a process to be spontaneous:         (A) The entropy of the system must increase;         (B) The energy of the system must decrease;         (C) The energy of the universe must decrease.         (D) The entropy of the universe must increase;

A student determines the value of the equilibrium constant to be 55.2 for the following reaction. C2H4(g) + H,O(g)- →CH;CH2OH(g) Based on this value of Keqi AG° for this reaction is expected to be (greater,less) than zero. Calculate the free energy change for the reaction of 1.64 moles of C,H.(g) at standard conditions at 298K. kJ rxn

Gibbs free energy (G�) is a measure of the spontaneity of a chemical reaction. It is the chemical potential for a reaction, and is minimized at equilibrium. It is defined as G=H−TS�=�−�� where H� is enthalpy, T� is temperature, and S� is entropy. The chemical reaction that causes magnesium to corrode in air is given by 2Mg+O2→2MgO2Mg+O2→2MgO in which at 298 K, ΔH∘rxn = −1204 kJΔ�rxn∘ = −1204 kJ and ΔS∘rxn = −217.1 J/KΔ�rxn∘ = −217.1 J/K.     Part A Part complete What is the standard Gibbs free energy for this reaction? Assume the commonly used standard reference temperature of 298 K.     Part B What is the Gibbs free energy for this reaction at 5958 KK ? Assume that ΔHΔ� and ΔSΔ� do not change with temperature. Express your answer to two decimal places and include the appropriate units.     Part C At what temperature Teq�eq do the forward and reverse corrosion reactions occur in equilibrium? Express your answer as an integer and include the appropriate units.

P2C.4 From the following data, determine A,H° for diborane, B,H,(g), at 298 K: (1) B,H,(g) + 30,(g) → B,O,(s) +3 H,O(g) A̟H° =-1941kJ mol -> %3D (2) 2B(s) + 를0,(g) → B,O, (s) A̟H® =-2368 kJ mol (3) H,(g)+±O,(g) → H,O(g) A,H° =-241.8kJ mol %3D