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Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
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- 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.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.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)A 1 mol quantity of hydrogen gas sample was heated at constant pressure from a temperature of 300 K to 500 K. With the entropy transformation process as a function of temperature variation, the heat capacity equation at constant pressure was estimated , Cp = 6.9469 - 0.199 x 10-3 T + 4.808 x 10-7 T2 (J K-1mol-1). Determine the entropy change of the process.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.
- 1 mol super cooled liquid water transformed to solid ice at -10 oC under 1 atm pressure. a) Calculate entropy change of the system, surrounding and universe. (temperature of the environment is -10 °C) b) Make some comments on entropy changes from the obtained data Please use the following data for water : Melting entalpy of ice (ΔHmelting) at 0°C and 1 bar is 6020 J mol-1. Cp (H2O (s)) = 37,7 J mol-1 K-1 Cp (H2O (l)) = 75,3 J mol-1 K-1The protein lysozyme unfolds at a transition temperature of 75.5 °C and the standard −1 enthalpy of transition is 509 kJmol . Calculate the entropy of unfolding of lysozyme at 25.0 °C, given that the difference in the constant pressure heat capacities upon unfolding is 6.28 kJK−1mol−1 and can be assumed to be independent of temperature. Hint: Imagine that the transition at 25.0 °C occurs in three steps: (i) heating of the folded protein from 25.0 °C to the transition temperature, (ii) unfolding at the transition temperature, and(iii) cooling of the unfolded protein to 25.0 °C. (Note: Because the entropy is a state function, the entropy change at 25.0 °C is equal to the sum of the entropy changes of the steps.)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, 130
- The temperature dependence of the heat capacity of non-metallic solids is found to follow the Debye T3 -law at very low temperatures, with Cp,m = aT3. (a) Derive an expression for the change in molar entropy on heating for such a solid. (b) For solid nitrogen, a= 6.15 x 10-3 J K-4 mol-1. What is the molar entropy of solid nitrogen at 5 K?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 AnswerP3B.7 A block of copper of mass 500 g and initially at 293K is in thermalcontact with an electric heater of resistance 1.00 kΩ and negligible mass. Acurrent of 1.00A is passed for 15.0 s. Calculate the change in entropy of thecopper, taking Cp,m = 24.4 JK−1mol−1. The experiment is then repeated with thecopper immersed in a stream of water that maintains the temperature of thecopper block at 293K. Calculate the change in entropy of the copper and thewater in this case.P3B.8 A block of copper (Cp,m = 24.44 JK−1mol−1) of mass 2.00 kg and at0 °C is introduced into an insulated container in which there is 1.00molH2O(g) at 100 °C and 1.00 atm. Assuming that all the vapour is condensed toliquid water, determine: (a) the final temperature of the system; (b) the heattransferred to the copper block; and (c) the entropy change of the water, thecopper block, and the total system. The data needed are given in ExerciseE3B.7a.