At a certain temperature, 4.0 mol NH3 is introduced into a 2.0 L container, and the NH3 partially dissociates by the reaction.
2 NH3(g) N2(g) + 3 H2(g)
At equilibrium, 2.0 mol NH3 remains. What is the value of K for this reaction?
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To find the value of K for this reaction, we need to set up the equilibrium expression and solve for K. The equilibrium expression for the given reaction is:
K = [N2] * [H2]^3 / [NH3]^2
Given that 2.0 mol NH3 remains at equilibrium, we can use this information to calculate the concentrations of N2 and H2:
[NH3] = (initial moles of NH3 - moles of NH3 at equilibrium) / volume of container
= (4.0 mol - 2.0 mol) / 2.0 L
= 2.0 mol / 2.0 L
= 1.0 M
[N2] = moles of N2 at equilibrium / volume of container
= 0 / 2.0 L
= 0 M
[H2] = moles of H2 at equilibrium / volume of container
= 0 / 2.0 L
= 0 M
Now we can substitute these concentrations into the equilibrium expression to solve for K:
K = [N2] * [H2]^3 / [NH3]^2
= (0 M) * (0 M)^3 / (1.0 M)^2
= 0 / 1.0
= 0
Therefore, the value of K for this reaction is 0.
To find the value of K for this reaction, we can use the equilibrium expression. The equilibrium expression for this reaction is:
K = [N2]^a * [H2]^b / [NH3]^c
Where [N2], [H2], and [NH3] represent the molar concentrations of N2, H2, and NH3 respectively at equilibrium, and a, b, and c represent the coefficients in the balanced equation. In this case, a = 1, b = 3, and c = 2.
Given that 2.0 mol NH3 remains at equilibrium, we can calculate the concentration of NH3 by using the ideal gas law:
PV = nRT
Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since we are given the number of moles (2.0 mol NH3) and the volume (2.0 L), we can rearrange the ideal gas law to solve for the pressure:
P = nRT / V
Substituting the values, we get:
P = (2.0 mol)(R)(T) / (2.0 L)
Now, we have the concentration of NH3 (which is equal to the pressure) in terms of the temperature.
Next, we can use stoichiometry to find the concentrations of N2 and H2. Since the reaction has a 1:3 ratio for NH3 to H2, and a 1:1 ratio for NH3 to N2, the concentration of N2 will be:
[N2] = c * [NH3] = 2.0 mol/L * P
And the concentration of H2 will be:
[H2] = b * [NH3] = 3 * 2.0 mol/L * P
Now, we can substitute these concentrations into the equilibrium expression for K:
K = [N2]^a * [H2]^b / [NH3]^c
K = (2.0 mol/L * P)^1 * (3 * 2.0 mol/L * P)^3 / (2.0 mol/L * P)^2
Simplifying the equation, we get:
K = (2.0 mol/L * P) * (6.0 mol/L * P)^3 / (2.0 mol/L * P)^2
K = (2.0 mol/L * P) * (216.0 mol^3/L^3 * P^3) / (4.0 mol^2/L^2 * P^2)
K = 432.0 mol^4/L^4 * P^4 / 8.0 mol^2/L^2 * P^2
K = 54.0 mol^2/L^2 * P^2
Therefore, the value of K for this reaction is equal to 54.0 mol^2/L^2 * P^2, where P is the pressure (concentration) of NH3 at equilibrium.