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What is the ∆H, ∆S, and ∆G when we isothermally mix 1.0 mol of toluene and 2.5 mol of benzene,
assuming ideal behavior?
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- Discuss the factors affecting the sign of ∆G in the relationship ∆G= ∆H - T∆S. Give examples for each of the situations.What is the equilibrium constant for a reaction that is carried out at 25 °C (298 K) with ∆H° = 20 kcal/mol and ∆S° = 5.0 * 10-2 kcal mol -1 K-1? What is the equilibrium constant for the same reaction carried out at 125 °C?What is the ΔH of the reaction KS→K+S?K+S→KS ΔH=+32.5 kJKS+K→K2S ΔH=+38.2 kJ A +32.5 kJ B +6 kJ C +70.7 kJ D -32.5 kJ
- Under base-catalyzed conditions, two molecules of acetone can condense to form diacetone alcohol. At room temperature (25 °C), about 5% of the acetone is converted to diacetone alcohol. Determine the value of ∆G° for this reaction.What is the ΔH of the reaction 2KS+2K→2K2S?K+S→KS ΔH=+32.5 kJKS+K→K2S ΔH=+38.2 kJ A +70.7 kJ B +65 kJ C +6 kJ D +76.4 kJ(a) Use the data given below and calculate ∆Ho, ∆So, ∆Go, and Kp at 25° C for the reaction:2 COCl2 (g) → 2 CO (g) + 2 Cl2 (g) (b) Calculate ∆G for the reaction at 250 °C. (c) At what temperature (°C) is ∆G equal to zero? In what temperature range is this reaction productfavored? Compound ∆ Ho, kJ/mol So, J/mol• KCOCl2 (g) -218.8 283.53CO (g) -110.52 197.67Cl2 (g) 0 223.07
- The standard free energy variation, at 25 ºC, for equilibrium: Glucose-6-phosphate (G-6-P) Glucose-1-phosphate (G-1-P) is ΔGº '= + 7280 J / mol. Calculate a) The equilibrium constant of the reaction. b) The real change in free energy when one mole of G-6-P is transformed into G-1-P, both concentrations remaining constant and equal to 10mM and 2 mM respectively.Consider the following equilibrium: N2O41g2 ∆ 2 NO21g2Thermodynamic data on these gases are given in Appendix C. You may assume that ∆H° and ∆S° do not vary with temperature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases?Calculate the ∆H° value for the following reaction:
- Use the following reactions to answer the next two questions. 4PbO (s) → 4Pb (s) + 2O2 (g) ∆H° = 876 kJ C (s) + O2 (g) → CO2 (g) ∆H° = -394 kJ 3PbCO3 (s) → 3PbO (s) + 3CO2 (g) ∆H° = 258 kJ A) Calculate ∆H° for the reaction: 2PbCO3 (s) → 2Pb (s) + 2C (s) + 3O2 (g) B)Calculate the enthalpy change, ∆H° (in kJ) that is involved in the production of 2.50 gram of solid carbonCalculate ΔH for the following reaction: CH4 (g) + O2 (g) ⇌ CO2 (g) + H2O (l) Compound ΔH CH4 (g) -74.8 kJ/mol H2O (l) -285.8 kJ/mol CO2 (g) -393.5 kJ/moThe equilibrium constant for the reaction Na2(g) → 2Na(g) is 2.47 at 1000. K. Calculate the value of ∆G∘ for this reaction under these conditions.