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The reaction below has reached equilibrium Which change will cause more HCl to form?
Kclark71
H2(g) + I2(g) 2HI(g), ΔH = +52 kJ/mol-Lowering the temperature.
Wiki User
∙ 15y ago
Increasing the concentration of Cl₂ will cause more HCl to form since it will shift the equilibrium towards the right to consume the excess Cl₂.
Wiki User
∙ 15y ago2Cl2(g) + 2H2O(g) 4HCl(g) + O2(g), ΔH = +115 kJ/mol-Decreasing the pressure of the system.
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The reaction shown below reaches equilibrium with the concentrations H2O2 equals 0.15 H2O equals 0.21 O2 equals 0.25 What is the equilibrium constant for this reaction?
The equilibrium constant (Kc) for the reaction would be [O2]/([H2O2]^[H2O]) = 0.25/(0.15*0.21) = 7.94
The reaction shown below reaches equilibrium with the concentrations H2O2 equals 0.37 H2O equals 0.24 O2 equals 0.92 What is the equilibrium constant for this reaction?
The equilibrium constant (Kc) for a reaction can be calculated using the concentrations of the products and reactants at equilibrium. In this case, Kc = [O2]/([H2O]^2). Plugging in the given values, Kc = (0.92)/((0.37)^2) ≈ 6.56.
The reaction below has reached equilibrium What will be the effect of adding CO2 to this system?
Adding CO2 to the system will shift the equilibrium to the right, favoring the formation of more products (H2 and CO) in order to relieve the stress caused by the increase in CO2 concentration. This shift will increase the concentrations of H2 and CO while decreasing the concentration of the reactant, CH4.
The equilibrium constant for the reaction below is 0.625. At equilibriumO 2 0.40 and H2O 0.20. What is the equilibrium concentration of H2O2?
Given the equilibrium constant (Kc) is 0.625 and the concentrations of O2 and H2O at equilibrium are 0.40 and 0.20 respectively, you can use the equilibrium expression Kc = [H2O2] / ([O2] * [H2O]) to solve for the equilibrium concentration of H2O2. Plugging in the values, you can calculate the concentration of H2O2 at equilibrium.
The equilibrium constant for the reaction below is 31.25 at equilibrium so2 0.03 o2 0.05 and h2s 0.15 what is the equilibrium concentration of H2O?
The equilibrium constant expression for the reaction is Kc = [H2O]^2/[SO2][O2]. Given the concentrations at equilibrium, we can solve for [H2O]. Plugging in the values, we get 31.25 = [H2O]^2 / (0.03)(0.05). Solving for [H2O] gives us [H2O] = sqrt(31.25 * 0.03 * 0.05), which is approximately 0.275M.
The equilibrium constant for the reaction below is 0.625 At equilibrium O2 0.40 and H2O 0.20 What is the equilibrium concentration of H2O2?
0.16
The equilibrium constant for the reaction below is 0.49 At equilibrium O2 equals 0.11 and N2 equals 0.15 What is the equilibrium concentration of NO?
0.09
The equilibrium constant for the reaction below is 31.25 at equilibrium so2 0.03 o2 0.05 and h2s 0.15 what is the equilibrium concentration of H2O?
The equilibrium constant expression for the reaction is Kc = [H2O]^2/[SO2][O2]. Given the concentrations at equilibrium, we can solve for [H2O]. Plugging in the values, we get 31.25 = [H2O]^2 / (0.03)(0.05). Solving for [H2O] gives us [H2O] = sqrt(31.25 * 0.03 * 0.05), which is approximately 0.275M.
The reaction shown below reaches equilibrium with the concentrations H2O2 equals 0.55 H2O equals 0.75 O2 equals 0.15 What is the equilibrium constant for this reaction?
0.28
The reaction shown below reaches equilibrium with the concentrations H2O2 equals 0.05 H2O equals 0.11 O2 equals 0.07 What is the equilibrium constant for this reaction?
0.34
The reaction shown below reaches equilibrium with the concentrations H2O2 equals 0.15 H2O equals 0.21 O2 equals 0.25 What is the equilibrium constant for this reaction?
The equilibrium constant (Kc) for the reaction would be [O2]/([H2O2]^[H2O]) = 0.25/(0.15*0.21) = 7.94
The reaction shown below reaches equilibrium with the concentrations H2O2 equals 0.37 H2O equals 0.24 O2 equals 0.92 What is the equilibrium constant for this reaction?
The equilibrium constant (Kc) for a reaction can be calculated using the concentrations of the products and reactants at equilibrium. In this case, Kc = [O2]/([H2O]^2). Plugging in the given values, Kc = (0.92)/((0.37)^2) ≈ 6.56.
The equilibrium constant for the reaction below is 0.625. At equilibriumO 2 0.40 and H2O 0.20. What is the equilibrium concentration of H2O2?
Given the equilibrium constant (Kc) is 0.625 and the concentrations of O2 and H2O at equilibrium are 0.40 and 0.20 respectively, you can use the equilibrium expression Kc = [H2O2] / ([O2] * [H2O]) to solve for the equilibrium concentration of H2O2. Plugging in the values, you can calculate the concentration of H2O2 at equilibrium.
The reaction below has reached equilibrium What will be the effect of adding CO2 to this system?
Adding CO2 to the system will shift the equilibrium to the right, favoring the formation of more products (H2 and CO) in order to relieve the stress caused by the increase in CO2 concentration. This shift will increase the concentrations of H2 and CO while decreasing the concentration of the reactant, CH4.
What is the equilibrium for the reaction below C(s)+o2(g) co2(g)?
96
The reaction shown below reaches equilibrium with the concentrations CO2 equals 0.15 CO equals 0.03 O2 equals 0.05 What is the equilibrium constant for this reaction?
H2(g) + I2(g) 2HI(g)18.6
The reaction below is at equilibrium What is the effect of removing fluorine gas from the system?
Br2(g) + 5F2(g) 2BrF5(g)-1.The reverse reaction rate is higher than the forward reaction rate. 2. BrF5 is consumed.
