Cantidades, constantes y cambios de condiciones de equilibrio
Introducido| English | Significado |
|---|---|
| dynamic equilibrium/daɪˈnæmɪk ˌiːkwɪˈlɪbrɪəm/ | equilibrio dinámico |
| equilibrium constant/ˌiːkwɪˈlɪbrɪəm ˈkɒnstənt/ | constante de equilibrio |
| partial pressure/ˈpɑːʃl ˈpreʃə/ | Presión parcial |
Why are equal reaction rates not equal concentrations?
- A closed reversible system can reach dynamic equilibrium 动态平衡 while containing much more product than reactant. Forward and reverse rates are equal; the amounts need not be.
- An equilibrium constant 平衡常数 describes the equilibrium composition for a stated reaction and temperature. First write the equation, phases and direction; changing the equation changes its expression.
Build the expression from the equation
- For $aA+bB\rightleftharpoons cC+dD$, use $K_c=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}$ with equilibrium concentrations. Coefficients become powers. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the expression is $\dfrac{[NH_3]^2}{[N_2][H_2]^3}$.
- In the usual heterogeneous expression, omit pure solids and pure liquids. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), $K_c=[CO_2]$ while both solids remain present. A dissolved solute is not omitted merely because it is aqueous. Derive units from the actual expression.
¿Cuál es la expresión Kc usual para CaCO₃(s) ⇌ CaO(s) + CO₂(g), mientras ambos sólidos están presentes?
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Los sólidos puros se omiten en la expresión usual, dejando [CO₂]. Cambiar la cantidad de un sólido no entra en Kc mientras permanezcan las fases requeridas.
Use an equilibrium mole inventory before dividing by volume
- For A(g) + B(g) ⇌ C(g), start with 0.500 mol each of A and B and no C in 2.00 dm³. If 0.200 mol C forms, the 1:1:1 ratio consumes 0.200 mol each reactant: equilibrium amounts are 0.300, 0.300 and 0.200 mol.
- Divide each amount by the same volume: $[A]=[B]=0.150\,\dfrac{\mathrm{mol}}{\mathrm{dm}^3}$ and $[C]=0.100\,\dfrac{\mathrm{mol}}{\mathrm{dm}^3}$. Thus $K_c=\dfrac{0.100}{0.150\times0.150}=4.44\,\dfrac{\mathrm{dm}^3}{\mathrm{mol}}$. Initial concentrations cannot be substituted for equilibrium concentrations.
The volume does not cancel in every expression. Here the numerator has one concentration factor and the denominator two, leaving a volume factor overall. By contrast, for A ⇌ B, Kc = [B]/[A] = n(B)/n(A) in the same vessel and is unitless in the course expression.
A(g) + B(g) ⇌ C(g): cantidades de equilibrio son 0.300, 0.300 y 0.200 mol en 2.00 dm³. Calcule Kc en dm³ por mol.
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Las concentraciones de equilibrio son 0.150, 0.150 y 0.100 mol por dm³. Kc = 0.100/(0.150 × 0.150) = 4.44 dm³ por mol. Usar solo cantidades omitiría el factor de volumen necesario.
Convert gas amounts to partial pressures
- Partial pressure 分压 is mole fraction × total pressure: $p_i=\dfrac{n_i}{n_{total}}P$. For N₂O₄(g) ⇌ 2NO₂(g), $K_p=\dfrac{p(NO_2)^2}{p(N_2O_4)}$. Use the specified pressure unit; this syllabus uses atm.
- At equilibrium, 1.00 mol N₂O₄ and 2.00 mol NO₂ have total pressure 3.00 atm. Their partial pressures are 1.00 and 2.00 atm, so $K_p=2.00^2/1.00=4.00\,\mathrm{atm}$. For the carbonate decomposition, $K_p=p(CO_2)$ while both pure solids are present.
N₂O₄(g) ⇌ 2NO₂(g): cantidades de equilibrio son 1.00 y 2.00 mol, con presión total 3.00 atm. Encuentre Kp en atm.
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Las fracciones molares 1/3 y 2/3 dan presiones parciales 1.00 y 2.00 atm. Kp = (2.00)²/1.00 = 4.00 atm.
Distinguish changing composition from changing the constant
- At fixed temperature, changing concentration or pressure can change equilibrium composition but not K. A catalyst speeds approach to equilibrium without changing K or its composition. Raising temperature lowers K for an exothermic forward reaction and raises K for an endothermic forward reaction; this change in K explains the temperature-dependent composition.
- A large K favours products for its stated expression; it does not prove rapid reaction or zero remaining reactant. Where the relationship $\Delta S_{total}=R\ln K$ is supplied, use the appropriately standardised dimensionless K: positive, zero and negative values correspond to K > 1, K = 1 and K < 1. Check the stated temperature and standard-state conventions before using it.
Do not take the logarithm of an unconverted dimensional Kp value. Do not insert total pressure in place of every gas partial pressure, include pure solid amounts in the usual expression, or claim a catalyst increases the equilibrium yield.
Write phases and coefficients, find equilibrium amounts, then calculate concentrations or partial pressures and derive K with its units. Temperature changes K; concentration, pressure and catalysts do not change its value at a fixed temperature.
A temperatura fija, añadir un catalizador no aumenta la constante de equilibrio.
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Ambas direcciones se vuelven más rápidas. El sistema alcanza la misma composición de equilibrio antes, con la misma K a temperatura fija.
Para una reacción directa endotérmica, ¿cómo cambia K al aumentar la temperatura?
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La dirección directa endotérmica se favorece al aumentar la temperatura, por lo que la expresión de equilibrio de producto a reactivo aumenta. Esto es un cambio en K, a diferencia de un cambio de presión a temperatura fija.