The natural tendency of any chemical reaction to proceed in a certain direction at constant temperat — Physical Chemistry — Thermodynamics Chemistry Question
In which direction does a chemical reaction proceed?
The natural tendency of any chemical reaction to proceed in a certain direction at constant temperature and pressure is determined by the sign of the Gibbs energy of the reaction, ∆G. This is the universal principle. If ∆G < 0, the reaction can proceed predominantly in the forward direction (a product-favored reaction). If ∆G > 0 the reaction can proceed predominantly in the reverse direction (a reactant-favored reaction). When ∆G = 0 the reaction is at equilibrium.
The standard reaction Gibbs energy, ∆G°, can be calculated from the tabulated Gibbs energies of formation of the reactants and products (see the Table).
Calculate the equilibrium constant of reaction (1) at 1627 °C. Can the reaction proceed predominantly in the forward direction if the initial partial pressure of O2 is below 1.00 Torr?
2 Ni(l) + O2(g) = 2 NiO(s) (1)
Model Answer
The standard Gibbs energy of the reaction (1) is equal to the Gibbs energy of formation of NiO, multiplied by two:
∆1900G° = 2 × (–72.1) = – 144.2 kJ mol–1
The equilibrium constant and the equilibrium partial pressure of oxygen at 1900 K are:
K = exp(–∆G° / RT) = exp(144200 / (8.314 × 1900)) = 9215,
p(O2) = 1 / K = 1.085 × 10–4 atm = 0.0825 Torr
If the oxygen pressure is above the equilibrium value, the reaction will proceed from the left to the right to reach the equilibrium state. So the answer is
0.0825 Torr < p(O2) < 1.00 Torr.
The standard Gibbs energy of the reaction
TiO2(s) + 3 C(s) = 2 CO(g) + TiC(s) (2)
is positive at 727 °C.
Calculate the equilibrium pressure of CO at 727 °C.
What should be the reaction conditions to allow for the forward reaction to be the predominant process at this temperature if this is possible at all?
Model Answer
The reaction proceeds forward as long as ∆G, not ∆G° is negative! The following equation is valid for the reaction (2):
∆G = ∆G° + RT ln p(CO)^2
(solid reactants and products are considered to be pure substances, they do not contribute to this equation). The reaction proceeds from the left to the right if ∆G < 0:
0 > ∆G° + RT ln p(CO)^2,
p(CO) < exp(–∆G° / 2RT)
Using the data from Table 1 we obtain:
∆G° = –162.6 + 2 × (–200.2) – (–757.8) = 194.8 kJ mol–1.
p(CO) < exp(–194800 / (2 × 8.314 × 1000)) = 8.17×10–6 atm.
Therefore, if the partial pressure of CO in the system is below 8.17×10–6 atm, the reaction can predominantly proceed from the left to the right.
Calculate the standard Gibbs energy of the reaction
3 H2 + N2 = 2 NH3 (3)
at 300 K. Can the forward reaction be the predominant process under the following conditions: p(NH3) = 1.0 atm, p(H2) = 0.50 atm, p(N2) = 3.0 atm?
In fact the reaction does not occur at 300 K at a noticeable rate. Why?
Model Answer
Using the data from Table 1, the following expression for ∆G of the reaction (3) is derived
∆G = ∆G° + RT ln(p(NH3)^2 / (p(H2)^3 p(N2))) = 2 × (–16260) + (8.314 × 300) × ln(1.0 / (0.50^3 × 3.0)) =
= –30100 J mol–1 = –30.1 kJ mol–1.
At 300 K the reaction (3) is allowed to proceed from the left to the right only.
However, formation of ammonia is extremely slow under these conditions due to the kinetic restrictions.