Optimizing Approximations in Equilibrium Problems
When to Neglect “x” in Equilibrium Problems
In the resolution of equilibrium problems, it is often necessary to neglect certain additive and subtractive terms in order to avoid lengthy and unnecessary calculations. We will examine equilibrium problems to see if and when such approximations can be made.
Equilibrium Problems
Consider the decomposition reaction of sulfur trioxide into sulfur dioxide and molecular oxygen:
2 SO3(g) ⇄ 2 SO2(g) + O2(g)
The equilibrium constant Kc is 1.6 ∙ 10-10 at 300 °C. We are given the initial concentration of SO3 as 0.100 M and asked to calculate the concentrations of the species at equilibrium. Utilizing an I.C.E. chart:
(State: Initial, Change, Equilibrium)
2 SO3(g) ⇄ 2 SO2(g) + O2(g)
Initial: 0.100
Change: -2x +2x +x
Equilibrium: 0.100 -2x 2x x
The expression for the equilibrium constant is:
Kc = 1.6 ∙ 10-10 = [SO2]2[O2]/[ SO3]2
Substituting the values obtained from the I.C.E. chart:
Kc = 1.6 ∙ 10-10 = (2x)2(x) / (0.100 – 2x)2 = 4×3 / (0.0100 + 4×2 – 0.4 x)
Solving this cubic equation directly is complex, hence the need for approximations. The subtractive term 2x can be ignored in the denominator. However, assuming x to be zero would make all variables zero, which is incorrect. Instead, the assumption is made for the subtractive term 2x relative to 0.100:
0.100 – 2x ≈ 0.100
This assumption is validated by the very small value of Kc, implying that a small amount of SO3 dissociates and can be considered negligible compared to 0.100.
From this assumption, the expression becomes:
Kc = 1.6 ∙ 10-10 = (2x)2(x) / (0.100)2 = 4×3/0.0100
Multiplying both sides by 0.0100:
1.6 ∙ 10-12 = 4×3
Dividing both sides by 4:
4.0 ∙ 10-13 = x3
Thus, x = ∛4.0 ∙ 10-13 = 7.4 ∙ 10-5
Substituting the value found for x:
2x = 2 (7.4 ∙ 10-5) = 1.5 ∙ 10-4 = [SO2]
x = 7.4 ∙ 10-5 = [O2]
0.100 – 2 (7.4 ∙ 10-5) = 0.100 = [SO3]
Validating the result by substituting the obtained values back into the equilibrium constant: Kc = (1.5 ∙ 10-4)2(7.4 ∙ 10-5)/0.100 = 1.6 ∙ 10-10, confirming the correctness of the assumption.
In the examined case, neglecting 2x in relation to 0.100 was justified due to the small value of Kc and the relatively high concentration of SO3. However, in general, an objective method is necessary to determine the validity of such an assumption. As a general rule, chemists assume x to be negligible if the subtractive or additive term is less than 5% relative to the initial concentration of the species. Hence, the best way to decide the validity of an assumption in a specific calculation is to test and then evaluate it against the rule.
For the reaction between nitrogen monoxide and oxygen at 200°C, with Kc at 3 ∙ 106, and initial concentrations of NO at 0.100 M and O2 at 0.050 M, the reaction is:
2 NO(g) + O2(g) ⇄ 2 NO2(g)
Using an I.C.E. chart to calculate the equilibrium concentrations.Calcolo dell’equilibrio di una reazione chimica
Nel processo di bilanciamento dell’equilibrio di una reazione chimica è necessario considerare attentamente i valori delle costanti di equilibrio ottenuti nel corso dell’analisi. Per ottenere risultati precisi, dobbiamo assumere che l’equilibrio sia spostato a destra senza trascurare alcun termine sottrattivo.
Per esprimere in termini numerici tale stato di equilibrio, è necessario considerare che tutto il monossido di azoto si sia trasformato in biossido di azoto e, successivamente, prendere in considerazione l’equilibrio.
Una volta ottenuti i valori pertinenti, sostituiamo tali valori nell’espressione della costante di equilibrio per trovare le concentrazioni delle specie in soluzione all’equilibrio. Questo calcolo ci permette di ottenere le concentrazioni di ciascuna specie chimica coinvolta nella reazione.
Il risultato di tali calcoli evidenzia che le concentrazioni delle specie all’equilibrio sono:
1. NO₂ = 0.098 M
2. NO = 0.0018 M
3. O₂ = 9.4 ∙ 10^-4 M
Il processo di bilanciamento dell’equilibrio chimico è essenziale per determinare le condizioni di una reazione e gli effetti dei vari parametri coinvolti.