At 2010K, the equilibrium constant Kc is 4*10^-4 for the following:
N2+O2-->2NO
If the equilibrium concentration of N2 is .28M and O2 is .38M, what is the equilibrium concentration of NO?
a)1.8*10^-9
b)2.1*10^-5
c)4.3*10^-5
d)6.5*10^-3
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Kc = (NO)^2/(N2)(O2)
You have only one unknown. Solve for that.
To find the equilibrium concentration of NO, we need to use the equilibrium constant expression and the given equilibrium concentrations of N2 and O2. The equilibrium constant expression for the given reaction is:
Kc = [NO]^2 / ([N2] * [O2])
In this equation, [NO] represents the equilibrium concentration of NO, [N2] represents the equilibrium concentration of N2, and [O2] represents the equilibrium concentration of O2.
Given equilibrium concentrations are:
[N2] = 0.28 M
[O2] = 0.38 M
Let's substitute these values into the equilibrium constant expression:
4*10^-4 = [NO]^2 / (0.28 M * 0.38 M)
Now, we can rearrange the equation to solve for [NO]:
[NO]^2 = (4 * 10^-4) * (0.28 M * 0.38 M)
[NO]^2 = 4.076 * 10^-5 M^2
Taking the square root of both sides gives us:
[NO] = √(4.076 * 10^-5 M^2)
[NO] ≈ 6.387 * 10^-3 M
Therefore, the equilibrium concentration of NO is approximately 6.387 * 10^-3 M.
Since none of the answer choices match exactly, we can round the value to the nearest answer choice, which is d) 6.5 * 10^-3.