Example The Haber process for the manufacture of ammonia is a classic example of
ID: 862799 • Letter: E
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Example The Haber process for the manufacture of ammonia is a classic example of the role of Le Chatelier's principle in an industrial process: Critical Thinking Questions Consider the Haber process shown in the example. a) Is this system at equilibrium at 25 degree C when (NH2) = 20.0 M. (N2) = 1.30 M, and (H2) = 0.010 M in a 1.00 L container? Explain your reasoning. b) The equilibrium system described above id disturbed by the addition of 0.40 moles of N2. Determine whether or not the system is still at equilibrium by calculating Q and comparing to K. What must happen at the molecular level to reestablish equilibrium? Predict how the equilibrium position will shift in response to each of the following stress (will the equilibrium shift toward the products and produce more NH3? Will the equilibrium shift toward the reactants and produce more N2 and H2? Or will the position of equilibrium remain unchanged?) Explain your reasoning for at least one of these answers by using concepts of Q and K.Explanation / Answer
a) Is this system at equilibrium at 25 degree C when
(NH3) = 20.0 M.
(N2) = 1.30 M, and
(H2) = 0.010 M in a 1.00 L container?
Explain your reasoning.
N2 + 3H2 = 2NH3
Keq = 1.8*10^8
Q = (NH3)^2/((N2)*(H2)^3) =
3.1*10^8
Q > Keq . So the system is not in equilibrium.
b)
The equilibrium system described above is disturbed by the addition of 0.40 moles of N2.
V = 1L
N2 = (1.3+0.4)/1 =1.7M
Q = (NH3)^2/((N2)*(H2)^3) =
2.35*10^8
Q > Keq.
So the system is not at equilibrium.
As Q > Keq, the reaction will proceed towards left. So, NH3 will decompose to form more N2 and H2 till Q = Keq.
C)
Concentration of N2 will not change Keq. Only Q will change.
3.
A) decrease H2
Keq = (NH3)eq^2/((N2)eq*(H2)eq^3)
Q = (NH3)^2/((N2)*(H2)^3)
H2 < H2eq
Q > Keq
So more reactant will be formed
B) increase H2
more product will be formed
C) decrease NH3
More product will be formed
D) increase NH3
More reactant will be formed.
E) decrease N2
More reactant will be formed
F)
increase N2
More product will be formed.
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