Calculate the value of the free energy change, ΔG, for the reaction below at 175.0ºC when the pressures of H2S (g) = 0.0100 atm, SO2(g) = 0.0250 atm, and H2O (g) = 2.50 atm. 2 H2S (g) + SO2 (g) → 3 S (s) + 2 H2O (g) ΔGº = –90.88 kJ ΔHº = –146.47 kJ ΔSº = –186.45 J/K
Calculate the value of the free energy change, ΔG, for the reaction below at 175.0ºC when...
Calculate the value of the free energy change, ΔG, for the reaction below at 227.0ºC when the pressures of NO (g) = 2.00 atm, O₂ (g) = 10.00 atm, and NO₂ (g) = 0.0250 atm. 2 NO (g) + O₂ (g) → 2 NO₂ (g) ΔGº = -70.54 kJ ΔHº = -114.14 kJ ΔSº = -146.43 J/K ΔG = ? kJ
(CM13201_S20), Calculate the value of the free energy change, AG, for the reaction below at 127.0°C when the pressures of NOCI (g) = 5.00 atm, NO (g) = 0.0200 atm, and Cl (g) - 0.0100 atm. 2 NOCI (9) - 2 NO (g) + Cl2 (g) AG° = 41.00 kj AH° = 77.08 kJ AS° = 121.24 J/K AGE Check
Given the reference thermodynamic data below taken at 25°C, calculate the value of the equilibrium constant for the reaction shown at 800.0ºC COCl2 (g) ⇄ CO (g) + Cl2 (g) ΔGº = 69.46 kJ ΔHº = 110.38 kJ ΔSº = 137.24 J/K K = Answer at 800.0ºC
Hydrogen cyanide is produced industrially from ammonia and methane at 1000ºC utilizing a platinum-rhodium catalyst by the following reaction: 2 NH3(g) + 3 O2(g) + 2 CH4(g) → 2 HCN(g) + 6 H2O(g) a) Calculate ΔGº, ΔHº, and ΔSº for this reaction at 25.00ºC. ΔGº ΔHº ΔSº b) Why is this reaction performed at such a high temperature (1000ºC)? _____________________________________________________________________________ _____________________________________________________________________________ c) Calculate the value of the equilibrium constant for this reaction at 25.00ºC. ___________________________________ d) What is the value...
Calculate the standard change in Gibbs free energy of the following reactions at Standard Ambient Temperature and Pressure (SATP where T = 25°C and P = 1 atm) and label them as spontaneous or nonspontaneous. (a) 2 SO2(g) + O2(g) 2 SO3(g): ΔΗ.-197.8 k , dS.-188.0 J/K kJ, and the reaction is Selectv (b) 2 C6H6(/) + 15 O2(g) 12 CO2(g) + 6 H2O(/); ΔΗ.-6535.0 kJ, S.-439.2 J/K kJ, and the reaction is Select (c) C(diamond); C(graphite); ΔΗ--19 ki, as...
Given the thermodynamic data below, calculate the value of the equilibrium constant for the reaction shown at 25.0ºC H₂ (g) + I₂ (g) ⇄ 2 HI (g) ΔHº = -9.48 kJ ΔSº = +21.79 J/K K = Answer at 25.0ºC
Using values of ΔG°f, calculate ΔG°rxn for the following reaction. Is the reaction product-favored or reactant-favored? SiCl4(g) + 2 Mg(s) → 2 MgCl2(s) + Si(s) ΔG°f (kJ/mol) HgS(s) -50.6 SO2(g) -300.13 H2S(g) -33.56 H2O(ℓ) -228.59 SiCl4(g) -622.76 MgCl2(s) -591.59 ____ kJ/mol? product-favored or reactant-favored?
Use the given data at 500 K to calculate ΔG°for the reaction 2H2S(g) + 3O2(g) → 2H2O(g) + 2SO2(g) Substance H2S(g) O2(g) H2O(g) SO2(g) ΔH°f(kJ/mol) -21 0 -242 -296.8 S°(J/K·mol) 206 205 189 248
Free-energy change, ΔG∘, is related to cell potential, E∘, by the equation ΔG∘=−nFE∘ where n is the number of moles of electrons transferred and F=96,500C/(mol e−) is the Faraday constant. When E∘ is measured in volts, ΔG∘ must be in joules since 1 J=1 C⋅V. Calculate the standard cell potential at 25 ∘C for the reaction X(s)+2Y+(aq)→X2+(aq)+2Y(s) where ΔH∘ = -679 kJ and ΔS∘ = -195 J/K .
Calculate Δ G for the reaction 2 SO2 (g) + O2 (g) → 2 SO3 (g) when P of SO2 = 0.500 atm, P of O2 = 0.0100 atm, and P of SO3 = 0.100 atm. The value for Δ G^o for this reaction at 298 K is -141.6 kJ. ΔG = ???? kJ I got -160.9 kj but it is incorrect.