At room temperature the equilibrium constant Kc for the reaction: 2NO(g)=N2(g)+O2(g) is 1.4x10^30. In the atmosphere at room temp. the concentration of nitrogen gas is 0.33 mol/L and the concentration of oxygen gas is about 25% that value. Calculate the equilibrium concentration of nitrogen monoxide in the atmosphere.
initial:
NO = 0
N2 = 0.33
O2 = 0.33 x 0.25 = ??
change:
NO = 2x
N2 = -x
O2 = -x
equilibrium:
NO = 2x
N2 = 0.33-x = 0.33 since x is so small.
O2 = 0.0825-x = 0.0825 same reason.
Substitute into Kc expression and solve for x.
Post your work if you get stuck.
To calculate the equilibrium concentration of nitrogen monoxide (NO) in the atmosphere, we need to set up an expression using the given information and the equilibrium constant (Kc) for the reaction.
The balanced equation for the reaction is:
2NO(g) ⇌ N2(g) + O2(g)
We know that the equilibrium constant (Kc) for the reaction is 1.4x10^30.
The given concentration of nitrogen gas (N2) is 0.33 mol/L. The concentration of oxygen gas (O2) is 25% of the concentration of nitrogen gas, i.e., 0.25 * 0.33 mol/L = 0.0825 mol/L.
Let's assign the equilibrium concentration of nitrogen monoxide (NO) as "x" mol/L.
Using the concentration values, we can set up the following expression for Kc:
Kc = [N2] * [O2] / [NO]^2
Substituting the given values:
1.4x10^30 = (0.33 mol/L) * (0.0825 mol/L) / (x mol/L)^2
Now, let's solve for the equilibrium concentration of nitrogen monoxide (NO):
1.4x10^30 = (0.033 mol^2) / x^2
x^2 = (0.033 mol^2) / 1.4x10^30
x = √[(0.033 mol^2) / 1.4x10^30]
x ≈ 1.94x10^-16 mol/L
Therefore, the equilibrium concentration of nitrogen monoxide (NO) in the atmosphere at room temperature is approximately 1.94x10^-16 mol/L.
To calculate the equilibrium concentration of nitrogen monoxide (NO) in the atmosphere, we need to use the equilibrium constant (Kc) and the given concentrations of nitrogen gas (N2) and oxygen gas (O2).
The equilibrium constant expression for the reaction is:
Kc = [N2][O2] / [NO]^2
In this case, we are given the value of Kc (1.4x10^30) and the concentrations of N2 and O2. Let's denote the equilibrium concentration of NO as x (in mol/L).
Given:
[N2] = 0.33 mol/L
[O2] = 0.25 * 0.33 mol/L (since the concentration of O2 is 25% of that of N2)
Substituting these values into the equilibrium constant expression, we get:
1.4x10^30 = (0.33)(0.25*0.33) / x^2
Simplifying the expression further:
1.4x10^30 = 0.27225 / x^2
To solve for x, let's rearrange the equation:
x^2 = 0.27225 / (1.4x10^30)
x^2 ≈ 1.94x10^(-32)
Taking the square root of both sides, we find:
x ≈ 4.41x10^(-17) mol/L
Therefore, the equilibrium concentration of nitrogen monoxide (NO) in the atmosphere at room temperature would be approximately 4.41x10^(-17) mol/L.