please help and explain.
At a certain temperature, Keq = 10.5 for the equilibrium below.
CO(g) + 2 H2(g) equilibrium reaction arrow CH3OH(g)
Calculate the following concentrations. (b) [H2] in an equilibrium mixture containing 1.01 mol/L CO and 0.335 mol/L CH3OH
At 2273 K, Keq = 6.2 multiplied by 10-4 for the following reaction.
N2(g) + O2(g) equilibrium reaction arrow 2 NO(g)
If [N2] = 0.04700 mol/L and [O2] = 0.01200 mol/L, what is the concentration of NO at equilibrium?
please answer and explain this one as well.
CO(g) + 2 H2(g) <==> CH3OH(g)
Keq = 10.5 = (CH3OH)/(CO)(H2)^2
You know K, (CH3OH), and (CO). Substitute and solve for (H2)
For the N2 + O2 ==> 2NO problem, is 0.04700 M and 0.01200 M the concentrations at the beginning of the reaction and you want to find the equilibrum concns for NO OR are those concns those at equilibrium and you want to find the equilibrium concn of NO?
To calculate the concentration of H2 in an equilibrium mixture, we can use the equilibrium constant expression and set it up using the given values.
The equilibrium constant expression for the given reaction is Keq = [CH3OH] / ([CO] * [H2]^2), where [CO], [H2], and [CH3OH] are the concentrations of the respective substances at equilibrium.
We are given the value of Keq (10.5) and the concentrations of CO (1.01 mol/L) and CH3OH (0.335 mol/L) in the equilibrium mixture. Let's assume the concentration of H2 is x mol/L.
So, the equilibrium constant expression can be rearranged as:
10.5 = 0.335 / (1.01 * x^2)
To solve for x, we can rearrange the equation:
x^2 = 0.335 / (1.01 * 10.5)
x^2 = 0.03152
x = √(0.03152)
By solving this calculation, we find that x ≈ 0.1777 mol/L. Therefore, the concentration of H2 in the equilibrium mixture is approximately 0.1777 mol/L.