Equilibrium concentrations: analyzing a real chemistry Unified State Exam task

Analysis of a Unified State Exam task on chemical equilibrium: calculation of the initial methane concentration and equilibrium oxygen concentration via reverse reaction analysis, taking reactant excess into account.

Take as a lesson

Concept of Chemical Equilibrium

Chemical equilibrium is the state of a reversible reaction in which the rate of the forward reaction equals the rate of the reverse reaction. At the beginning of the process, the reactant concentration is at its maximum and products are absent, so the forward reaction rate is high, while the reverse rate is zero. As products accumulate, the forward reaction rate decreases and the reverse rate increases. At the point where they equalize, dynamic equilibrium is established: the reaction continues in both directions, but no visible changes in concentrations occur.

It is important to understand that the equilibrium concentration of a substance is not the amount formed in the forward reaction, but what remains after the partial return of products to reactants via the reverse reaction.

Task Condition

Methane and oxygen were placed in a constant-volume reactor. The reaction proceeds:

CH₄ + 2O₂ ⇌ 2CO + 4H₂ + ... (simplified scheme for illustrating the calculation using 1 : 2 : 2 : 4 coefficients).

Typical Error: Calculation Based on Oxygen

It is tempting to calculate using the known oxygen concentration (0.5 mol/L). According to stoichiometry, 0.5 mol/L of O₂ yields 0.5 mol/L of CO and 1.0 mol/L of H₂. Then the reverse reaction "consumes" 0.6 mol/L of CO and 1.2 mol/L of H₂, forming 0.6 mol/L of CH₄. However, this contradicts the condition: the equilibrium methane concentration must be 0.2 mol/L.

The reason for the discrepancy is excess and limiting reactant. The condition does not state which reactant is in excess. If the calculation is performed using the excess substance, stoichiometry distorts the result. Therefore, calculations must be based on the limiting reactant — in this case, methane.

Correct Solution Steps

  1. Analysis of the reverse reaction via methane. The equilibrium concentration of CH₄ is 0.2 mol/L — this is the amount regenerated by the reverse reaction.
  2. Stoichiometric recalculation. 0.2 mol/L of CH₄ were formed from 0.2 mol/L of CO and 0.4 mol/L of H₂ (coefficients 1 : 2 : 4 → for methane 1, carbon monoxide 2, hydrogen 4; the reverse pathway yields a proportion of 0.2 : 0.2 (per coefficients in the initial reaction 2:2... — accounted for by the ratio from the lecture)).
  3. Products actually formed in the forward reaction: CO — 0.4 + 0.2 = 0.6 mol/L; H₂ — 0.8 + 0.4 = 1.2 mol/L.
  4. Initial methane concentration: according to stoichiometry, CH₄ → 2CO, meaning 0.6 mol/L of methane reacted. Adding the equilibrium residue of 0.2: x = 0.6 + 0.2 = 0.8 mol/L (or another option depending on the reaction notation; in the lecture, x = 0.6 was used as the amount that reacted in the forward reaction).

Calculation of Equilibrium Oxygen Concentration

Checking the excess: for 0.6 mol/L of reacted methane, only 0.3 mol/L of O₂ is required (coefficient 2 balanced by the task ratio). Given 0.5 mol/L — oxygen is indeed in excess, confirming the hypothesis.

The reverse reaction returns oxygen: 0.2 mol/L of CO in the reverse reaction yield 0.1 mol/L of O₂. The regenerated oxygen is added to the excess remainder (0.2 mol/L):

y = 0.2 + 0.1 = 0.3 mol/L.

Methodological Conclusions

Key points

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