Jumlah, konstanta kesetimbangan, dan perubahan kondisi
Diperkenalkan| English | Makna |
|---|---|
| dynamic equilibrium/daɪˈnæmɪk ˌiːkwɪˈlɪbrɪəm/ | keseimbangan dinamik |
| equilibrium constant/ˌiːkwɪˈlɪbrɪəm ˈkɒnstənt/ | konstanta kesetimbangan |
| partial pressure/ˈpɑːʃl ˈpreʃə/ | tekanan parsial |
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.
Manakah ekspresi Kc umum untuk CaCO₃(s) ⇌ CaO(s) + CO₂(g), ketika kedua zat padat hadir?
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Zat padat murni diabaikan dalam ekspresi umum, menyisakan [CO₂]. Mengubah jumlah zat padat tidak masuk dalam Kc selama fase yang diperlukan tetap ada.
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): jumlah kesetimbangan adalah 0.300, 0.300 dan 0.200 mol dalam 2.00 dm³. Hitunglah Kc dalam dm³ per mol.
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Konsentrasi kesetimbangan adalah 0.150, 0.150 dan 0.100 mol per dm³. Kc = 0.100/(0.150 × 0.150) = 4.44 dm³ per mol. Menggunakan jumlah mol saja akan mengabaikan faktor volume yang diperlukan.
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): jumlah kesetimbangan adalah 1.00 dan 2.00 mol, dengan total tekanan 3.00 atm. Temukan Kp dalam atm.
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Fraksi mol 1/3 dan 2/3 menghasilkan tekanan parsial 1.00 dan 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.
Pada suhu tetap, penambahan katalis meningkatkan konstanta kesetimbangan.
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Kedua arah menjadi lebih cepat. Sistem mencapai komposisi kesetimbangan yang sama lebih cepat, dengan K yang sama pada suhu tetap.
Untuk reaksi endotermik maju, bagaimana kenaikan suhu mengubah K?
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Reaksi endotermik maju difavoritkan oleh kenaikan suhu, sehingga ekspresi kesetimbangan produk-terhadap-pereaksi meningkat. Ini merupakan perubahan K, berbeda dengan perubahan tekanan pada suhu tetap.