steps and procedures oT Of calculating the pressure of 3.169 KPa as a result of mixing up the two tanks named 1&2 (linked by a valve) with the environment temperature of 25 °C. ake the following given data. (i) 400
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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 Hm=6000Jmol-1Ethanoic acid (acetic acid) was evaporated in container of volume 21.45 cm3 at 437 K and at an external pressure of 101.9 kPa, and the container was then sealed. The mass of acid present in the sealed container was 0.0519 g. The experiment was repeated with the same container but at 471 K, and it was found that 0.0380 g of the acid was present. Calculate the equilibrium constant for the dimerization of the acid in the vapour, and the standard enthalpy of the dimerization reaction.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…
- Suppose now that argon is added to the mixture in the previous exercise to bring the composition closer to real air, with mole fractions 0.780. 0.210, and 0.0096, respectively. (a) What is the additional change in molar Gibbs energy and entropy at 298 K? (b) Is the mixing spontaneous?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)What is the value of the equilibrium constant when ΔGorxn = 25 kJ·mol-1 and T=298 K?
- A sample containing 2.50 moles of He (1 bar, 350. K) is mixed with 1.75 mol of Ne (1 bar, 350. K) and 1.50 mol of Ar (1 bar, 350. K). Calculate AGmixing and ASmixing.Calculate (a) the (molar) Gibbs energy of mixing, (b) the(molar) entropy of mixing when the two major componentsof air (nitrogen and oxygen) are mixed to form air at 298 K.The mole fractions of N2 and 0 2 are 0.78 and 0.22. respectively. (c) Is the mixing spontaneous?Prove that the equation below holds for an ideal gas mixture (Partial Molar Gibbs Energy of a Perfect gas Gi=Gio +RTlnYi )
- (a) The heat evolved upon dissolution of atmospheric ozone into water is determined to be +5 kcal/mol at 298 K. Given that ozone has a Henry’s constant of 9.4 × 10−3 mol/L/atm at 298 K, determine its Henry’s constant at 273 K. Note that the definition of Henry’s constant is for the process air to water. (b) If the atmospheric ozone concentration in a polluted air is 100 ppbv, what is the concentration in equilibrium with a body of water in that region at 298 K? Note that you need to assume no losses of ozone other than absorption into water. Compatible Polymers A and B are to be blended to achieve a glass-transition temperatureof 60 oC. Tg’s of Polymers A and B are 30 and 85 oC, respectively, using an appropriateequation determine the composition of the blend.At 0 °C liquid water and ice are in equilibrium. What is the effect on the difference in molar Gibbs energy when the pressure is increased from 1.0 bar to 100 bar? The mass densities of the two phases at 0 °c are ρ(ice) = 0.9150 g cm-3 and ρ(liquid) = 0.9999 g cm-3.