At 2273 K the equilibrium constant for the reaction
2 NO(g) « N2(g) + O2(g) KC = 2400
What would be the equilibrium concentration of N2 if the initial concentration of NO was 0.1778 M?
Do you mean for this reaction be be written as
N2 + O2 ==> 2NO with Kc = 2400. That's the way you have your arrows.
To find the equilibrium concentration of N2, we need to use the equilibrium constant (Kc) expression and calculate it using the initial concentration of NO (nitric oxide).
The balanced equation for the reaction is:
2 NO(g) ⇌ N2(g) + O2(g)
The equilibrium constant expression is:
Kc = [N2] * [O2] / [NO]^2
Given:
Initial concentration of NO (NO) = 0.1778 M
Equilibrium constant (Kc) = 2400
Let's assume the equilibrium concentration of N2 is x M.
Using the equilibrium constant expression, we can set up the equation:
Kc = [N2] * [O2] / [NO]^2
Substituting the values:
2400 = (x) * ([O2]) / (0.1778)^2
Since the stoichiometric coefficient of N2 is 1, the equilibrium concentration of N2 will also be x M. So we can rewrite the equation as:
2400 = (x) * ([O2]) / (0.1778)^2
Simplifying:
2400 = (x) * ([O2]) / 0.03167
Now, we need to determine the concentration of O2 at equilibrium. Since the stoichiometric coefficient of O2 is also 1, the equilibrium concentration of O2 will be x M.
Substituting x for [O2]:
2400 = (x) * (x) / 0.03167
Rearranging the equation:
2400 * 0.03167 = x^2
Solving for x using a calculator:
x^2 = 76.0848
Taking the square root:
x ≈ 8.72
Therefore, the equilibrium concentration of N2 is approximately 8.72 M when the initial concentration of NO is 0.1778 M.
To find the equilibrium concentration of N2, we need to use the equilibrium constant expression and the initial concentration of NO.
The equilibrium constant expression for the given reaction is:
KC = [N2] * [O2] / [NO]^2
Where [N2], [O2], and [NO] are the equilibrium concentrations of N2, O2, and NO, respectively.
Given that KC = 2400, we can substitute this value into the equation:
2400 = [N2] * [O2] / [NO]^2
Since the reaction produces 1 mole of N2 for every 2 moles of NO, we can assume that the equilibrium concentration of N2 is 2x (where x is the equilibrium concentration of NO).
Substituting this into the equilibrium constant expression:
2400 = (2x) * [O2] / (x)^2
Simplifying the equation:
2400 = 2 * [O2] / x
Now we can use the initial concentration of NO to calculate the equilibrium concentration of N2.
Given that the initial concentration of NO is 0.1778 M, we substitute this value for x:
2400 = 2 * [O2] / 0.1778^2
Now we can solve for [O2]:
[O2] = 2400 * 0.1778^2 / 2
Calculating this:
[O2] ≈ 47.860 M
Therefore, the equilibrium concentration of N2 is approximately (2 * 0.1778) M, or 0.3556 M.