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- The molar enthalpy of fusion of ice at 273.15 K and one atm is ΔfusHm (H2O)=6.01 kJ mol-1, andthe molar entropy of fusion under the same conditions is ΔfusSm (H2O)=22.0 J K-1 mol-1. Show that(a) ΔfusGm (H2O)=0 at 273.15 K and one atm, (b) ΔfusG,m (H2O) < 0 when the temperature is greaterthan 273.15 K, and (c) ΔfusGm (H2O) > 0 when the temperature is less than 273.15 K.Assume that solutions of ethylbenzene : benzene behave ideally. a) Calculate the entropy of mixing if 40 g of ethylbenzene is mixed into 50g of benzene.b) At room temperature (298 K), what is ΔmixG for mixing 40g of ethylbenzene (PhEt) and 50g of benzene (PhH)?c) Would you notice a temperature change associated with the process in parts a) & b)?d) Instead, you mix 40g of benzyl alcohol (PhMeOH) into 50g of benzene (PhH). Let’s denote the difference between this process and the process in part b) as: ΔΔG = ΔmixG [PhMeOH ∶ PhH] − ΔmixG [PhEt ∶ PhH] What do you expect the sign of ΔΔG to be? (ΔΔG<0, ΔΔG≈0, or ΔΔG>0)Briefly justify your answer.Calculate the standard reacton entropy at 298K of the following: 1. CH3CHO (g) + O2 -----→ CH3COOH (l) 2. Sucrose [C12H22O11(s)] + O2(g) ----→ CO2 (g) + H2O (l) 3. Zn(s) + Cu2+ (aq) ---→ Zn2+ (aq) + Cu (s)
- Assume that solutions of ethylbenzene : benzene behave ideally. a) Calculate the entropy of mixing if 40 g of ethylbenzene is mixed into 50 g of benzene.b) At room temperature (298 K), what is ΔmixG for mixing 40 g of ethylbenzene (PhEt) and 50 g of benzene (PhH)?c) Would you notice a temperature change associated with the process in parts a) & b)?d) Instead you mix 40 g of benzyl alcohol (PhMeOH) into 50 g of benzene (PhH). Let’s denote the difference between this process and the process in part b) as:ΔΔG = ΔmixG[ PhMeOH ∶ PhH ] − ΔmixG[ PhEt ∶ PhH ]What do you expect the sign of ΔΔG to be? ( ΔΔG < 0, ΔΔG ≈ 0, or ΔΔG > 0 )Briefly justify your answer.If the change of entropy of the surrounding increased by 0.27 KJ/mol.K due t oa reactiob carried out at a 25C tempreture ,which one of these following terms apply to the reaction?cricle all the correct answers. Exergonic endergonic exothermic endothermicThe change of heat capacity with temperature for solid aluminum is as follows:Cp(J/K mol) = 20.67 + 12.38 x10-3T (in the range of 25 -700°C )If S°298 = 28.35 J / K mol, what will be the entropy of aluminum at 700°C?
- For the complete neutralization reaction between a 1 M strong acid and a 1 M strong base of equal initial volumes that is marked by a temperature rise, the entropy of neutralization is positive (ΔSrxn > 0) whereas the Gibbs free energy is negative (ΔGrxn < 0). Is this true or false? Please explain.When nitric acid is produced industrially, nitrogen monoxide, NO, is first formed at high temperature. Bakefetr reacts NO on cooling further with oxygen to nitrogen dioxide: 2 NO(g) + O2 ⇌ 2 NO2 (g) Table 1: Thermodynamic data at 25°C. Bond ΔfHom Som Cop,m NO(g) 90.25 210.76 29.34 O2(g) 0.00 205.14 29.36 NO2(g) 33.18 240.06 37.20 1) Calculate (with all relevant intermediate calculations) the standard reaction Gibbs free energy, ΔrG25o, for reaction (1) at 25°C from the data in Table 1 2) Calculate (with all relevant intermediate calculations) the equilibrium constant K25, for reaction (1) at 25°C. 3) Industrially, however, the reaction does not proceed at 25°C but at 500°C. Therefore, calculate (with all relevant intermediate calculations) the standard reaction Gibbs free energy, ΔrG500o, for reaction (1) at 500°C under the assumption that the standard molar heat capacities, Cop, in Table 1 are independent of temperature in the interval [25°C, 500°C]Benzene (C6H6) has a melting point of 5.50C and an enthalpy of fusion of 10.04kJmol-1 at 25.00C. The molar heat capacities at constant pressure for solid and liquid benzene are 100.4JK-1mol-1 and 133.0JK-1mol-1, respectively. Calculate the change of entropy of system and change of entropy of the surrounding at 100C for the reaction of C6H6(l)---->C6H6(s)
- Part b-- Standard entropy = 205 J/mol KA sample of K(s) of mass 3.226 gg undergoes combustion in a constant volume calorimeter at 298.15 K. The calorimeter constant is 1849 J⋅K−Mol-, and the measured temperature rise in the inner water bath containing 1538 gg of water is 1.776 K. CP,m(H2O,l)=75.3J⋅mol−1⋅K−1CP,m(H2O,l)=75.3J⋅mol−1⋅K−1. 1) Calculate ΔU���f for K2O 2) Calculate ΔH∘f for K2OThe molar entropy of argon (Ar) is given by Sm = 36.4 + 20.8 ln T (in SI units). Calculate the change in Gibbs energy when 1.25 moles of argon is heated at constant pressure from 25 °C to 65 °C.