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
Calculate the value of the free energy change, ΔG, for the reaction below at 227.0ºC when...
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 150.0ºC when the pressures of H2S (g) = 0.0200 atm, SO2(g) = 0.400 atm, H2O (g) = 5.00 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 ΔG = Answer 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...
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
The equilibrium constant of a system, K, can be related to the standard free energy change, ΔG∘ ΔG∘=−RTlnK where T is a specified temperature in kelvins (usually 298 KK) and R is equal to 8.314 J/(K⋅mol) Under conditions other than standard state, the following equation applies: ΔG=ΔG∘+RTlnQ In this equation, Q is the reaction quotient and is defined the same manner as KK except that the concentrations or pressures used are not necessarily the equilibrium values. Part A Acetylene, C2H2,...
1. For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction 2NO(g)+O2(g)↽−−⇀2NO2(g) the standard change in Gibbs free energy is Δ?°=−69.0 kJ/mol . What is ΔG for this reaction at 298 K when the partial pressures are ?NO=0.500 atm , ?O2=0.400 atm , and ?NO2=0.900 atm ? 2. Given the following information A+B⟶2D Δ?∘=656.0 kJ Δ?∘=291.0 J/K C⟶D ΔH°=467.0 kJ ΔS°=−116.0 J/K calculate ΔG° at 298 K for...
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 .
21A. Calculate the standard free energy change, AGºat 298 K for the reaction 2COXg) + 2NO(g) -2CO(g) + N:(g) The standard free energy of formation for CO is - 137 kJ, for NO it is 87.6 kJ/mol and for CO, it is -394 kJ/mol. B. Calculate the free energy change. AG. at 298 K. given that the partial pressure of CO is 5.0 atm, that of NO is 4.0 atm, that of CO, is 3.0 atm & that of N,...