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 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 1018
- 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 = 37.4Given 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)Acetaldehyde (CH3CHO) is an important chemical both industrially and biologically. For instance, it is a (somewhat toxic) intermediate in the body's metabolism of ethanol into acetic acid, and thus is possibly implicated in the "hungover" symptoms of someone who has had too much to drink the night before. In aqueous solution, it establishes an equilibrium with a hydrated form, shown below. CH3CHO (aq) + H2O (l) <--> CH3CH(OH)2 (aq) You start with an aqueous sample, already at equilibrium, with the CH3CH(OH)2 (the hydrated form) at a concentration of 2.60 M. You have no information about how much, if any, of the anhydrous form (CH3CHO) is initially in the flask. If you add 2.0 M of CH3CHO to the reaction flask, and as the equilibrium is being restored the amount of CH3CH(OH)2 changes by 1.13 M, what is the final amount of CH3CHO?
- From the following equilibrium reactions: 2SO2(g) + O2(g) → 2SO3(g) K1 2C0(g) + O2(g) → 2C02(g) K2 Determine the equilibrium constant, K, for 2SO2(g) + 2CO2(g) 2SO3(g) + 2CO(g) O K=Kq/K2 O K=K1 + 1/K2 O K=K2/K1 O K=K¼×K2The following equilibria were measured at 823 K:CoO(s) + H2(g) ⇌ Co(s) + H2O(g) Kc = 67H2(g) + CO2(g) ⇌ CO(g) + H2O(g) Kc = 0.14 (a) Use these equilibria to calculate the equilibrium constant,Kc, for the reaction CoO(s) + CO(g) ⇌ Co(s)+ CO2(g) at 823 K. (b) Based on your answer to part(a), would you say that carbon monoxide is a stronger orweaker reducing agent than H2 at T = 823 K? (c) If youwere to place 5.00 g of CoO(s) in a sealed tube with a volumeof 250 mL that contains CO(g) at a pressure of 1.00atm and a temperature of 298 K, what is the concentrationof the CO gas? Assume there is no reaction at thistemperature and that the CO behaves as an ideal gas (youcan neglect the volume of the solid). (d) If the reactionvessel from part (c) is heated to 823 K and allowed tocome to equilibrium, how much CoO(s) remains?At 25 °C, the following reactions have the equilibrium constants noted to the right of their equations. 2CO(g) + O₂(g) ⇒ 2CO₂(g) Kç= 3.3 × 10⁹1 2H₂(g) + O₂(g) → 2H₂O(g) Kc= 9.1 × 108⁰ Use these data to calculate Kc for the reaction H₂O(g) + CO(g) ⇒ CO₂(g) + H₂(g) Kc= i
- Calculate the value of the equilibrium constant for the following system, given the data shown: H2 (g) + CO2 (g) ⇌ H2O(g) + CO(g) Concentrations at equilibrium: [H2] = 1.5 M [CO2] = 2.5 M [H2O] = 0.5 M [CO] = 3.0 MCalculate a value for the equilibrium constant for the reaction (6)*O = (6)0 + (6) ²0 given hv NO2 (9) NO(g) + O(g) K = 2.3 x 10-49 O3 (9) + NO(9) = NO,(g) + O2(9) K = 8.4 x 10-34 (Hint: When reactions are added together, the equilibrium expressions are multiplied.) K =Given the folowing equilibria: cog) + 2 H2(g) = CH;OH(g) H20(g) + C(s) = H2(g) + Co(g) K= 10. x 10-5 K= 6.0 x 10-5 What is the equilibrium constant for the following reaction? C(s) + H2(g) + H,0(g) = CH;OH(g) K = ? A. 1.6 x 104 В. 6.0 x 10-5 C. 1.6 x 10-9 D. 6.0 x 10-10 Е. 6.0 x 10-9