Use the provided information to calculate Kc for the following reaction at 550 °C: H2(g) + CO2(g) ⇌ CO(g) + H2O(g) Kc = ?CoO(s) + CO(g) ⇌ Co(s) + CO2(g) Kc1 = 490CoO(s) + H2(g) ⇌ Co(s) + H2O(g) Kc2 = 67
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Use the provided information to calculate Kc for the following reaction at 550 °C:
H2(g) + CO2(g) ⇌ CO(g) + H2O(g) Kc = ?
CoO(s) + CO(g) ⇌ Co(s) + CO2(g) Kc1 = 490
CoO(s) + H2(g) ⇌ Co(s) + H2O(g) Kc2 = 67
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- Consider the following reaction at equilibrium: CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) If CO(g) is removed, how will K(eq) change?The following equilibria were attained at 823 K:CoO(s)+H2(g)⇌Co(s)+H2O(g) Kc=67CoO(s)+CO(g)⇌Co(s)+CO2(g) Kc=480 Based on these equilibria, calculate the equilibrium constant for H2(g)+CO2(g)⇌CO(g)+H2O(g) at 823 K. Express your answer using two significant figures.Given the equilibrium constant valuesN2(g)+1/2O2(g)⇌N2O(g) KC=2.7×10−18 N2O4(g)⇌2NO2(g) KC=4.6×10−3 1/2N2(g)+O2(g)⇌NO2(g) KC=4.1×10−9 Determine a value of KC for the reaction: 2N2O(g)+3O2(g)⇌2N2O4(g)
- Given the following reactions and equilibrium constants: (1) HCN(g) + OH* — CN- + H₂O (g) K₁ =4.9 x 104 (2) H₂O(g) H¹(g) + OH- K₂ = 1.0 x 10-14 - What will be the equilibrium constant for: HCN(g) → CN- + H* (g) ? um constant 5.9 x 104 4.9 x 10-10 5.0 x 10-10 4.9 x 10-40 4.9 x 1018The equilibrium constant, Kc, for the following reaction is 18.1 at 773 K. 2NH3 (9) ⇒ N₂(g) + 3H₂(g) Calculate Ke at this temperature for the following reaction: 1/2N2(g) + 3/2H₂ (9) ⇒ NH3 (9) Kc =Given these reactions and their equilibrium constants, CO(g) + 3 H₂(g) → CH4(g) + H₂O(g) K = 3.93 N₂(g) + H₂(g) = NH3(g) Kc = 0.324 determine Ke for the following reaction. CO(g) + 2 NH3(g) CH4(g) + H₂O(g) + N₂(g) Your Answer: Kc = 37.4
- Express the equilibrium constant for the following reaction: 2 CH2Cl2 (g) + Cl2 (g) ⇌2 CHCl3 + H2 (g) k=?If the equilibrium constants for the two reactions 2 HCl(g) H2(g) + Cl2(g) K1 = 4.36x10-2 and I2(g) + Cl2(g) 2 ICl(g) K2 = 8.92x10-6 are denoted K1 and K2 respectively, then the equilibrium constant, K3, for the reaction below is 2 HCl(g) + I2(g) 2 ICl(g) + H2(g) K3= ?Use the information below to calculate ΔG°298 for the following reaction: NO (g) + O (g) ⇌ NO2 (g) ΔG°298 = ? 2 O3 (g) ⇌ 3 O2 (g) ΔG°298 = +489.6 kJ O2 (g) ⇌ 2 O (g) ΔG°298 =+463.4kJ NO (g) + O3 (g) ⇌ NO2 (g) + O2 (g) ΔG°298 = -199.5 kJ What does the value you calculated for ΔG°298 in Question 1 tell you about the reaction NO (g) + O (g) ⇌ NO2 (g) at this temperature?
- Consider the following reversible reaction at equilibrium: C6H12O6(aq) + 6 O2(g) 6 CO2(g) + 6 H2O(l). Given that this reaction is exothermic, if heat is added to the equilibrium system, how is the stress relieved?Given the following equilibrium constants at 427°C: O₂(g) + 4Na(1) 2Na₂O(s) O₂(g) + 2Na(1) 2NaO(g) O₂(g) + 2Na(1) ⇒ Na₂O₂(s) O₂(g) + Na(1) NaO₂ (s) 2.00×10-¹2 K₁ = 2.50x1049 You are correct. Your receipt no. is 153-9292 ? K₂ = 2.50×10⁹ What would be the value of the equilibrium constant for each of the following reactions, at 427°C? K3 = 2.00×1028 2Na₂O(s) +O₂(g) = 2NaO(g) + Na₂O₂(s) Submit Answer Tries 0/99 K4 = 3.33x1013 Previous Tries NaO(g) + NaO₂(s) Na₂O(s) + O₂(g)At 25°C, the equilibrium constant for the reaction 6 CIO3F(g) 2 CIF(g) + 4 ClO(g) + 7 O2(g) + 2 F2(g) is 32.6. Calculate the equilibrium constant at 25°C for the reaction }CIF(g) + ?CIO(g) + ¿O2(g) + }F2(8) 2 clo;F{g)