What is ΔG°(in kJmol-1) for 2 SO2(g) + O2(g)→2 SO3 (g) at 700 K, under standard conditions of 1 atm partial pressure for all gases, given that Kp = 3.0 × 104 at 700 K?.
A)+60
B)26
C)0.0
D)+26
E)60
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What is ΔG°(in kJmol-1) for 2 SO2(g) + O2(g)→2 SO3 (g) at 700 K, under standard...
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.
For the reaction 2 SO2(g) + O2(g) →→2 SO3(g) AG° = -140.3 kJ and AS = -187.9 J/K at 306 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 306 K. The standard enthalpy change for the reaction of 1.79 moles of so,(g) at this temperature would be
Calculate AG 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 AGº for this reaction at 298 K is -141.6 kJ. AG= O * kJ Is this reaction spontaneous or non-spontaneous under these conditions? Spontaneous or non-spontaneous? spontaneous
For the equilibrium: 2 SO3(g) < = > O2(g) + 2 SO2(g) Kp = 0.269 at 625 oC What is Kc at this temperature? Kp = Kc[RT]Δn R = 0.08206 L-atm/mol K
2) For the equilibrium: 2 SO3(g) <=> O2(g) + 2 SO2(g) Kp = 0.269 at 625 °C What is Ke at this temperature? Kp = Kc[RT]An R = 0.08206 L-atm/mol K
Use the provided thermodynamic data at 298 K and partial pressures given to decide in which direction (forward or backward) the reaction will proceed under these conditions: Compound ΔG (kJ/mol) Partial Pressure (atm) SO2 (g) -300.2 1.0 x 10-4 O2 (g) 0.0 0.20 SO3 (g) -371.1 0.10
The equilibrium constant, Kp, for the following reaction is 2.74 at 1.15x103K. 2803(g) 22502(g) + O2(g) + If an equilibrium mixture of the three gases in a 10.9 L container at 1.15*10²K contains SO3 at a pressure of 1.77 atm and SO2 at a pressure of 0.926 atm, the equilibrium partial pressure of O2 is atm.
04 Question (1 point) A reaction vessel contains an equilibrium mixture of SO2, O2, and SO3. The reaction proceeds such that: 2802(g) +0,(8) — 2803(e) 6th attempt The partial pressures at equilibrium are: Pso, = 0.001915 atm Po, = 0.001111 atm Pso, = 0.0166 atm Calculate Kp for the reaction. x 104 - 3 OF 16 QUESTIONS COMPLETED < 04/16 > + VIEW SOLUTION
Given the reaction: 2 SO2(g) + O2(g) ⇌ 2 SO3(g) Starting with 0.100 M of SO2 and 0.050 M of O2, which value of K should be used and why? a. KC because the chemicals are in the gas phase. b. KP because the concentrations are in Molarity. c. KP because the chemicals are in the gas phase. d. KC because the concentrations are in Molarity.
The equilibrium constant in terms of pressures, Kp, for the reaction of SO2 and O2 to form SO3 is 0.365 at 1.15×103 K: SO2(g) + O2(g) = 2SO3(g) A sample of SO3 is introduced into an evacuated container at 298 K and allowed to dissociate until its partial pressure reaches an equilibrium value of 0.867 atm. Calculate the equilibrium partial pressures of SO2 and O2 in the container. PSO2 = PO2 =