Le Chatelier’s principle states that «Every system in the state of equilibrium when subjected to a p — Analytical Chemistry Chemistry Question
Le Chatelier’s principle
Le Chatelier’s principle states that «Every system in the state of equilibrium when subjected to a perturbation responds in a way that tends to eliminate the effect» (P.W. Atkins “Physical Chemistry”).
Let us see how this principle works. Let a chemical equilibrium be established in the following reaction between the ideal gases:
3 H2 + N2 = 2 NH3 (1)
At the temperature of T = 400 K partial pressures of reactants and product are respectively: p(H2) = 0.376 bar, p(N2) = 0.125 bar, p(NH3) = 0.499 bar.
The equilibrium was disturbed. Let this disturbance be:
a) increase of the total pressure in the system at constant temperature,
b) increase of the amount of NH3 in the system at constant total pressure and temperature,
c) small increase of the amount of N2 in the system at constant total pressure and temperature,
d) small increase of the amount of H2 in the system at constant total pressure and temperature.
Calculate the standard Gibbs energy for the reaction (1) at T = 400 K.
Model Answer
∆G° = – RT ln K = – RT ln(p(NH3)^2 / (p(H2)^3 p(N2))) (2)
∆G° = – 8.314 × 400 × ln(0.499^2 / (0.376^3 × 0.125)) = –12100 J mol–1 = –12.1 kJ mol–1.
Write down the expression for the Gibbs energy of reaction (1) for any pressure of reactants and product after perturbation. This expression is called the isotherm of chemical reaction.
Model Answer
After perturbation, the Gibbs energy of the reaction is:
∆G = ∆G° + RT ln(p'(NH3)^2 / (p'(H2)^3 p'(N2))) (3)
The apostrophe ‘ denotes the partial pressures at the non-equilibrium state. The sign of ∆G (positive or negative) determines the direction in which the equilibrium shifts after perturbation.
Using the equation of isotherm from question 7.2 determine in which direction the reaction (1) will predominantly proceed after the disturbance of equilibrium as indicated in (a) – (d).
Model Answer
7.3, 7.4
Let us determine the sign of ∆G in all the considered cases. From equations (2) and (3), we get:
∆G = RT (2 ln(p'(NH3)/p(NH3)) – 3 ln(p'(H2)/p(H2)) – ln(p'(N2)/p(N2))) (4)
Reactants and product are ideal gases, so we can use the Dalton law. Molar fractions x can be calculated from the partial pressures:
x(NH3) = p(NH3)/P, x(H2) = p(H2)/P, x(N2) = p(N2)/P (5)
P is the total pressure in the system. Taking into account (5), equation (4) can be written in a form:
∆G = RT (2 ln(x'(NH3)/x(NH3)) – 3 ln(x'(H2)/x(H2)) – ln(x'(N2)/x(N2)) – 2 ln(P'/P)) (6)
In the case (a), only the last term in the right hand side of the equation (6) is non-zero. Since the total pressure is increased P′ > P, the right side of equation (6) is negative, ∆G < 0. The increase of the total pressure will push the reaction towards formation of additional amounts of ammonia. The reaction will proceed predominantly in the forward direction (a product-favored reaction).
In the case (b), only the last term on the right side of (6) is equal to zero. Molar fraction of ammonia increases, whereas molar fractions of hydrogen and nitrogen decrease: ln(x'(NH3)/x(NH3)) > 0, ln(x'(H2)/x(H2)) < 0, ln(x'(N2)/x(N2)) < 0. The right side of (6) is positive and ∆G > 0. In the case b), the reaction will proceed predominantly in the reverse direction towards formation of additional amounts of reactants.
In the case (c) similarly as in the case (b), all the molar fractions change after the addition of hydrogen to the system. After simple rearrangements of the equation (6) one gets:
∆G = RT (–3 ln(n'(H2)/n(H2)) – 2 ln((n(H2)+n(N2)+n(NH3))/(n'(H2)+n'(N2)+n'(NH3)))) (7)
where n is the number of moles of reactants or product. The first term in the right side of (7) is negative (n'(H2) > n(H2)) while the second one is positive.
Let us solve the inequality ∆G < 0:
–3 ln(n'(H2)/n(H2)) – 2 ln((n(H2)+n(N2)+n(NH3))/(n'(H2)+n'(N2)+n'(NH3))) < 0 (8)
Let n'(H2) = n(H2) + ∆H2, where ∆H2 is the number of moles of hydrogen added to the system. Since ∆H2 is small, n'(H2) ≈ n(H2). The inequality (8) can be written in the form:
(1 + ∆H2/n(H2))^3 (1 + ∆H2/(n(H2)+n(N2)+n(NH3)))^-2 > 1
Terms with the second and third powers of ∆H2 can be neglected, then:
3(∆H2/n(H2)) – 2(∆H2/(n(H2)+n(N2)+n(NH3))) > 0, or x(H2) < 3/2
This inequality is always valid, since molar fractions are less than one. It means that in the case (c) ∆G < 0, no matter what the initial composition of the mixture was. After addition of a small amount of hydrogen to the system the reaction will proceed predominantly in the direction of ammonia synthesis.
In the case (d) both hydrogen and nitrogen are reactants. Their roles in the reaction (1) are similar. It is reasonable to expect that in cases (c) and (d) the answer to the problem will be the same. However, let us look at equation (9) which is similar to equation (8):
∆G = RT (–ln(n'(N2)/n(N2)) – 2 ln((n(H2)+n(N2)+n(NH3))/(n'(H2)+n'(N2)+n'(NH3)))) (9)
In the right side of (9) the first term is negative (n'(N2) > n(N2)), while the second is positive.
Let us solve the inequality ∆G < 0:
–ln(n'(N2)/n(N2)) – 2 ln((n(H2)+n(N2)+n(NH3))/(n'(H2)+n'(N2)+n'(NH3))) < 0 (10)
Denote n'(N2) = n(N2) + ∆N2, then
(1 + ∆N2/n(N2))(1 + ∆N2/(n(H2)+n(N2)+n(NH3)))^-2 > 1
Again, term with the second power of ∆N2 can be neglected, and then:
1(∆N2/n(N2)) – 2(∆N2/(n(H2)+n(N2)+n(NH3))) > 0, thus, x(N2) < 1/2
Will the answers to question 3 change, if the initial equilibrium partial pressures in the system are: p(H2) = 0.111 bar, p(N2) = 0.700 bar, p(NH3) = 0.189 bar? Assume that temperature and total pressure in the system are the same as in questions 7.1 – 7.3.
Model Answer
If the molar fraction of nitrogen in the initial equilibrium mixture is less than 0.5 (question 7.3), the small increase of the amount of nitrogen will push the reaction towards the formation of ammonia. But if x(N2) > 1/2 (question 7.4) after the addition of nitrogen the reaction will proceed predominantly in the reverse direction towards formation of the reactants.
Thus, in some cases addition of the reactant can lead to the opposite results. This “strange conclusion” is in full accord with the Le Chatelier’s principle!