Calculate the change in S when one mole of water is heated from 263 to 283 K given the molar capacities in J.K-1 , Cp(ice) = 2.09 + 0.126T, Cp(water)=75.3, and change in Hm=6000Jmol-1
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Calculate the change in S when one mole of water is heated from 263 to 283 K given the molar capacities in
J.K-1 , Cp(ice) = 2.09 + 0.126T, Cp(water)=75.3, and change in Hm=6000Jmol-1
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- Calculate the change in S when one mole of water is heated from 263 to 283 K given the molar capacities inJ.K-1 , Cp(ice) = 2.09 + 0.126T, Cp(water)=75.3, and change in enthalpy at formation is 6000Jmol-1. P=101.32 kPaiv) Determine the volume change ΔV for the transformation from rhombic to monoclinic sulphur.v) Estimate the pressure at which rhombic and monoclinic sulphur coexist in equilibrium at a temperature of 373 K.Coiled coils are protein domains that lead to the formation of multimers (dimers,trimers, tetramers, etc.). They are found in a wide range of proteins. Here we willconsider the trimerization reaction. We may write the chemical equilibrium as3 M(aq) ⇄ T(aq)where M and T represent, respectively, the monomer and trimer species. Given that∆rG0 = −25.00 kJ mol−1 at 37 0C (with the standard state the normal one for reactionsin solution, namely c0 = 1.0 mol dm−3):A) What is the equilibrium constant for the formation of the trimer at 37 0C? B) Suppose that in a particular experiment at 37.00 0C the initial concentrations ofthe monomer if [M]0 = 2.00×10−2 µM and the initial concentration of the trimeris also [T]0 = 2.00 × 10−2 µM. What is ∆rG for the formation of the trimer atthe specified temperature? C) For the conditions given in part (B) will trimers spontaneously convert to monomers?Give a very short justification of your answer. D) Suppose that ∆rH0 and ∆rS0 are independent of the…
- The change in Gibbs energy that accompanies the combustion of C6H12O6(s) to carbon dioxide and water vapour at 25 °c is -2828 kJ mol-1. The potential energy of an object of mass m at a height h, relative to that at the Earth's surface is given by mgh, where g = 9.81 m s-2 is the acceleration of freefall. How much glucose does a person of mass 65 kg need to consume to climb through 10m?Calculate (a) the (molar) Gibbs energy of mixing, (b) the (molar) entropy of mixing when the two major components of air (nitrogen and oxygen) are mixed to form air at 298 K. The mole fractions of N2 and O2 are 0.78 and 0.22, respectively. (c) Is the mixing spontaneous?The standard Gibbs energy of formation of rhombic sulfur is zero, and that of monoclinic sulfur is +0.33 kJ mol−1 at 25 °C. The standard molar entropy of rhombic sulfur is 31.80 J K−1 mol−1, and that of monoclinic sulfur is 32.6 J K−1 mol−1. At what temperature will the transition occur at 1 bar? _______ K. 3 sig. fig.
- A sample of water vapour at 200 °c is compressed isothermally from 350 cm3 to 120 cm3. What is the change in its molar Gibbs energy?Suppose a certain small bird has a mass of 30 g. What is the minimum mass of glucose that it must consume to fly to a branch 10 m above the ground? The change in Gibbs energy that accompanies the oxidation of 1.0 mol C6H12O6(s) to carbon dioxide and water vapour at 25 °C is -2828 kJ.Ratio of specific heats of CO2 at 825K?
- The pressure dependence of G is quite different for gases and condensed phases. Calculate ΔGm for the process (C,solid,graphite,1bar,298.15K)→(C,solid,graphite,300.bar,298.15K) The density for graphite is 2250 kg⋅m−3 Calculate ΔGm for the process (He,g,1bar,298.15K)→(He,g,300.bar,298.15K)Derive a formula for the molar Gibbs free energy change of a closed system at constant temperature if the system obeys the equation of state PV = nRT + nCP where C is a constant.Without carrying out an explicit calculation, explain there lative values of the standard molar entropies (at 298 K) of the following substances: (a) Ne(g) (146 J K-1 mol-1) compared with Xe(g) (170 J K-1 mol-1), (b) H2O(g) (189 J K-1 mol-1) compared with D2O(g) (198 J K-1 mol-1), (c) C(diamond) (2.4 J K-1 mol-1) compared with C(g raphite) (5.7 J K-1 mol-1).