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- calculate the standard entropy change for the following reqction at 25*C. round to 4 significant figures. 2S(s,rhombic)+ 3O2(g) --> 2SO3(g)Use Entropy values from the Thermodynamic Data Table in Resource Section of Atkin's Physical Chemistry to calculate ΔS° (in J/K) for the combustion of propane.C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) Answer in 4 significant figures.Solid iodine, I2(s), has an enthalpy of sublimation ΔHsubl = 62.4 kJ mol-1. Suppose 5.36 g of iodine at a constant temperature of 35.0 oC is allowed to sublime into the atmosphere at a temperature of 50.8 oC. 1.Calculate the entropy change of the system (i.e. the iodine) in J K-1. Enter your answer into the first answer field in accordance with the question statement
- Consider the reaction:2C2H6(g) + 7O2(g)4CO2(g) ---------> 6H2O(g)Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.69 moles of C2H6(g) react at standard conditions. S°system = _______ J/KAssume 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.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 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.Using the thermodynamic information in the ALEKS Data tab, calculate the standard reaction entropy of the following chemical reaction: →+Al2O3s3H2g+2Als3H2Og Round your answer to zero decimal places. H2 0 , 0, 130When 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)Consider the reaction:1. 2SO2(g) + O2(g)2SO3(g)Using standard absolute entropies at 298K, calculate the entropy change for the system when 2.06 moles of SO2(g) react at standard conditions. S°system = J/K Submit Answer 2. Consider the reactionFe2O3(s) + 3H2(g)2Fe(s) + 3H2O(g)Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.80 moles of Fe2O3(s) react at standard conditions. S°surroundings = J/K Submit AnswerA 1.0 mol of Cu metal at T = 90°C is dropped into a calorimeter containing 3.5 moles of H2O at T = 25°C. The calorimeter is sealed to the outside environment with a negligible heat capacity. The heat capacity of liquid water is 75.0 J/Kmol and for Cu is 25.0 J/Kmol. a) What is the entropy change of the Cu? b) What is the entropy change of the water?