The gaseous substances A2 and B2 were mixed in a molar ratio 2 : 1 in a closed vessel at a temperatu — Analytical Chemistry Chemistry Question
Simple equilibrium
The gaseous substances A2 and B2 were mixed in a molar ratio 2 : 1 in a closed vessel at a temperature T1. When the equilibrium A2(g) + B2(g) = 2AB(g) was established the number of heteronuclear molecules in a gas phase became equal to the total number of homonuclear molecules.
Determine the equilibrium constant K1 for the above reaction.
Model Answer
The initial ratio A2 : B2 = 2 : 1
2 1 A2 + B2 = 2 AB x x 2x 2–x 1–x 2x
n(AB) = 2x = n(A2) + n(B2) = (2–x) + (1–x),
x = 0.75
K1 = n(AB)^2 / (n(A2) * n(B2)) = 1.5^2 / (1.25 * 0.25) = 7.2
Find the ratio of heteronuclear to homonuclear molecules at equilibrium if the substances are mixed in a ratio 1:1 at the temperature T1.
Model Answer
The initial ratio A2 : B2 = 1 : 1
1 1 A2 + B2 = 2AB y y 2y 1–y 1–y 2y
K1 = (2y)^2 / ((1-y) * (1-y))
The ratio of heteronuclear to homonuclear molecules:
n(AB) / (n(A2) + n(B2)) = 2y / ((1-y) + (1-y)) = y / (1-y) = 1.34
The equilibrium mixture obtained from the initial mixture A2 : B2 = 2 : 1 was heated so that equilibrium constant became K2 = K1 / 2.
How much substance B2 (in percent to the initial amount) should be added to the vessel in order to keep the same equilibrium amounts of A2 and AB as at the temperature T1?
Model Answer
New equilibrium constant: K2 = K1 / 2 = 3.6.
Equilibrium amounts: n(AB) = 1.5 mol, n(A2) = 1.25 mol, n(B2) = 0.25+x mol,
1.5^2 / (1.25 * (0.25+x)) = 3.6,
x = 0.25 mol = 25 % of initial amount of B2 should be added.
Consider the reaction yield η = neq(AB) / nmax(AB) as a function of the initial molar ratio A2 : B2 = x : 1 at any fixed temperature (nmax is the maximum amount calculated from the reaction equation). Answer the following questions qualitatively, without exact equilibrium calculations.
At what x the yield is extremal (minimal or maximal)?
Model Answer
Consider two initial mixtures: A2 : B2 = x : 1 and A2 : B2 = 1/x : 1 = 1 : x. It is clear that in both cases the equilibrium yield is the same, hence η(x) = η(1/x). The value x = 1 for such functions is the extremum point. We can prove it in the following way. Consider the identity:
near the point x = 1. If η(x) is an increasing or decreasing function at x = 1, then near this point both sides of the identity will have opposite signs. Hence, either η(x) = const (which is chemical nonsense), or x = 1 is the point of extremum.
What is the yield at: a) x → ∞; b) x → 0?
Model Answer
a) At x→∞ the very large amount of A2 will almost completely shift the equilibrium A2 + B2 = 2AB to the right, and almost all B2 will be converted to AB, the yield will tend to 1, η(x→∞) → 1.
b) At x→ 0 (1/x→∞) the situation is the same as in (a) if we interchange A2 and B2, that is η(x→0) → 1.
Draw the graph of η(x).
Model Answer
It follows from question 5 that at x = 1 the function η(x) has a minimum, because at x = 0 or x = ∞ it approaches the maximum possible value of 1. Qualitatively, the graph is as follows:
Now, consider the variable ratio A2 : B2 = x : 1 at a fixed total pressure.
At what x the equilibrium amount of AB is maximal?
Model Answer
Suppose we have in total 1 mol of A2 and B2, and the molar ratio A2 : B2 = x : 1. Then, the initial amounts of reagents are: n(A2) = x/(x+1), n(B2) = 1/(x+1). It follows from the symmetry between A2 and B2 that the equilibrium amount of AB will be the same for the molar ratios x and 1/x, hence x = 1 corresponds to the maximum or minimum neq(AB).
If x is very large (small), then the initial amount of B2 (A2) will be small and so will be neq(AB). Therefore, the maximum amount of AB will be obtained at A2 : B2 = 1 : 1.
The equilibrium calculation for this case is as follows.
0.5 0.5 A2 + B2 = 2 AB y y 2y 0.5–y 0.5–y 2y
K = (2y)^2 / (0.5 - y)^2
y = 0.5 * √K / (2 + √K)
n(AB) = 2y