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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
What happens if you lower the pressure of the system The system below is at equilibrium. 2co(g) plus o2(g) 2co2(g)?
Lowering the pressure of the system will cause the reaction to shift towards the side with fewer gas molecules to increase the pressure. In this case, the reaction will shift to the right, producing more CO2 gas molecules until a new equilibrium is reached.
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?
CO(g) + H2O(g) CO2(g) + H2(g)-The reverse reaction rate will be higher than the forward reaction rate until equilibrium is reached again.
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 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 0.55 H2O equals 0.75 O2 equals 0.15 What is the equilibrium constant for this reaction?
0.28
What is the equilibrium for the reaction below C(s)+o2(g) co2(g)?
96
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
What happens if you lower the pressure of the system The system below is at equilibrium. 2co(g) plus o2(g) 2co2(g)?
Lowering the pressure of the system will cause the reaction to shift towards the side with fewer gas molecules to increase the pressure. In this case, the reaction will shift to the right, producing more CO2 gas molecules until a new equilibrium is reached.
How would the graph below change if a catalyst were added to the reaction apex?
If a catalyst were added to the reaction apex, the graph would show a decrease in the activation energy barrier, allowing the reaction to proceed more quickly. However, the overall shape of the graph would remain the same, as a catalyst does not change the reactants or products, only the rate at which equilibrium is reached. Consequently, the transition state would occur lower on the energy axis, but the starting and final energy levels would stay constant.
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?
CO(g) + H2O(g) CO2(g) + H2(g)-The reverse reaction rate will be higher than the forward reaction rate until equilibrium is reached again.
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 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 0.49 At equilibrium O2 equals 0.11 and N2 equals 0.15 What is the equilibrium concentration of NO?
To find the equilibrium concentration of NO, first calculate the equilibrium constant expression using the given concentrations of O2 and N2. Then, rearrange the equilibrium constant expression to solve for the concentration of NO. Finally, substitute the values of O2 and N2 concentrations into the rearranged expression to find the equilibrium concentration of NO.
