Equilibrium amounts, constants and changing conditions
Введено| English | Значение |
|---|---|
| dynamic equilibrium/daɪˈnæmɪk ˌiːkwɪˈlɪbrɪəm/ | динамическое равновесие |
| equilibrium constant/ˌiːkwɪˈlɪbrɪəm ˈkɒnstənt/ | константа равновесия |
| partial pressure/ˈpɑːʃl ˈpreʃə/ | парциальное давление |
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.
Which is the usual Kc expression for CaCO₃(s) ⇌ CaO(s) + CO₂(g), while both solids are present?
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Pure solids are omitted in the usual expression, leaving [CO₂]. Changing a solid amount does not enter Kc while the required phases remain.
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): equilibrium amounts are 0.300, 0.300 and 0.200 mol in 2.00 dm³. Calculate Kc in dm³ per mol.
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Equilibrium concentrations are 0.150, 0.150 and 0.100 mol per dm³. Kc = 0.100/(0.150 × 0.150) = 4.44 dm³ per mol. Using amounts alone would omit the required volume factor.
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): equilibrium amounts are 1.00 and 2.00 mol, with total pressure 3.00 atm. Find Kp in atm.
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Mole fractions 1/3 and 2/3 give partial pressures 1.00 and 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.
At fixed temperature, adding a catalyst increases the equilibrium constant.
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Both directions become faster. The system reaches the same equilibrium composition sooner, with the same K at fixed temperature.
For an endothermic forward reaction, how does raising temperature change K?
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The endothermic forward direction is favoured by raising temperature, so the product-to-reactant equilibrium expression increases. This is a change in K, unlike a pressure change at fixed temperature.