Artikel sebelumnya telah menurunkan rumus orde reaksi untuk dua pereaksi menggunakan Aturan Cramer pada sistem 2×2. Artikel ini memperluas penurunan tersebut ke kasus tiga pereaksi, dengan semua konsentrasi boleh berbeda di setiap percobaan. Pola penurunannya persis sama hanya dimensi sistemnya yang bertambah dari 2×2 menjadi 3×3.
1. Persamaan Laju dan Linearisasi
Untuk reaksi dengan tiga pereaksi:
\[ v = k[A]^m[B]^n[C]^p \]
Ambil logaritma kedua ruas:
\[ \log v = \log k + m\log[A] + n\log[B] + p\log[C] \]
Persamaan ini linear terhadap tiga bilangan yang tidak diketahui: \(m\), \(n\), dan \(p\).
2. Butuh Berapa Percobaan?
Untuk dua pereaksi, ada dua bilangan tidak diketahui (\(m\) dan \(n\)), sehingga dibutuhkan minimal 3 percobaan, satu sebagai acuan, dua lainnya untuk membentuk sistem 2 persamaan setelah \(\log k\) dieliminasi.
Dengan tiga pereaksi, ada tiga bilangan tidak diketahui (\(m\), \(n\), dan \(p\)), sehingga dibutuhkan minimal 4 percobaan, satu sebagai acuan, tiga lainnya untuk membentuk sistem 3 persamaan.
3. Membangun Sistem Persamaan
| Perc. | [A] (mol/L) |
[B] (mol/L) |
[C] (mol/L) |
v (mol L-1 s-1) |
|---|---|---|---|---|
| 1 | [A]1 | [B]1 | [C]1 | v1 |
| 2 | [A]2 | [B]2 | [C]2 | v2 |
| 3 | [A]3 | [B]3 | [C]3 | v3 |
| 4 | [A]4 | [B]4 | [C]4 | v4 |
\[ \log v_1 = \log k + m\log[A]_1 + n\log[B]_1 + p\log[C]_1 \quad \cdots(1) \]
\[ \log v_2 = \log k + m\log[A]_2 + n\log[B]_2 + p\log[C]_2 \quad \cdots(2) \]
\[ \log v_3 = \log k + m\log[A]_3 + n\log[B]_3 + p\log[C]_3 \quad \cdots(3) \]
\[ \log v_4 = \log k + m\log[A]_4 + n\log[B]_4 + p\log[C]_4 \quad \cdots(4) \]
Kurangkan persamaan (1) dari (2), (3), dan (4):
\[ \log\frac{v_2}{v_1} = m\log\frac{[A]_2}{[A]_1} + n\log\frac{[B]_2}{[B]_1} + p\log\frac{[C]_2}{[C]_1} \quad \cdots(\text{I}) \]
\[ \log\frac{v_3}{v_1} = m\log\frac{[A]_3}{[A]_1} + n\log\frac{[B]_3}{[B]_1} + p\log\frac{[C]_3}{[C]_1} \quad \cdots(\text{II}) \]
\[ \log\frac{v_4}{v_1} = m\log\frac{[A]_4}{[A]_1} + n\log\frac{[B]_4}{[B]_1} + p\log\frac{[C]_4}{[C]_1} \quad \cdots(\text{III}) \]
\(\log k\) telah hilang. Tersisa sistem tiga persamaan dengan tiga bilangan tidak diketahui.
4. Sistem dalam Bentuk Matriks
Definisikan notasi ringkas:
\[ P = \log\frac{v_2}{v_1},\quad Q = \log\frac{v_3}{v_1},\quad R = \log\frac{v_4}{v_1} \]
\[ \alpha_1 = \log\frac{[A]_2}{[A]_1},\quad \alpha_2 = \log\frac{[A]_3}{[A]_1},\quad \alpha_3 = \log\frac{[A]_4}{[A]_1} \]
\[ \beta_1 = \log\frac{[B]_2}{[B]_1},\quad \beta_2 = \log\frac{[B]_3}{[B]_1},\quad \beta_3 = \log\frac{[B]_4}{[B]_1} \]
\[ \gamma_1 = \log\frac{[C]_2}{[C]_1},\quad \gamma_2 = \log\frac{[C]_3}{[C]_1},\quad \gamma_3 = \log\frac{[C]_4}{[C]_1} \]
Sistem persamaan (I), (II), (III) menjadi:
\[ \begin{pmatrix} \alpha_1 & \beta_1 & \gamma_1 \\ \alpha_2 & \beta_2 & \gamma_2 \\ \alpha_3 & \beta_3 & \gamma_3 \end{pmatrix} \begin{pmatrix} m \\ n \\ p \end{pmatrix} = \begin{pmatrix} P \\ Q \\ R \end{pmatrix} \]
5. Penerapan Aturan Cramer 3×3
Untuk sistem \(A\mathbf{x} = \mathbf{b}\) berukuran 3×3, Aturan Cramer menyatakan:
\[ m = \frac{\det(A_m)}{\det(A)}, \quad n = \frac{\det(A_n)}{\det(A)}, \quad p = \frac{\det(A_p)}{\det(A)} \]
di mana \(A_m\), \(A_n\), \(A_p\) adalah matriks \(A\) dengan kolom 1, 2, dan 3 diganti oleh vektor \(\mathbf{b}\) secara bergantian.
Determinan matriks 3×3 dihitung dengan ekspansi kofaktor baris pertama:
\[ D = \det(A) = \begin{vmatrix} \alpha_1 & \beta_1 & \gamma_1 \\ \alpha_2 & \beta_2 & \gamma_2 \\ \alpha_3 & \beta_3 & \gamma_3 \end{vmatrix} \]
\[ D = \alpha_1(\beta_2\gamma_3 - \gamma_2\beta_3) - \beta_1(\alpha_2\gamma_3 - \gamma_2\alpha_3) + \gamma_1(\alpha_2\beta_3 - \beta_2\alpha_3) \]
\[ D_m = \begin{vmatrix} P & \beta_1 & \gamma_1 \\ Q & \beta_2 & \gamma_2 \\ R & \beta_3 & \gamma_3 \end{vmatrix} = P(\beta_2\gamma_3 - \gamma_2\beta_3) - \beta_1(Q\gamma_3 - \gamma_2 R) + \gamma_1(Q\beta_3 - \beta_2 R) \]
\[ D_n = \begin{vmatrix} \alpha_1 & P & \gamma_1 \\ \alpha_2 & Q & \gamma_2 \\ \alpha_3 & R & \gamma_3 \end{vmatrix} = \alpha_1(Q\gamma_3 - \gamma_2 R) - P(\alpha_2\gamma_3 - \gamma_2\alpha_3) + \gamma_1(\alpha_2 R - Q\alpha_3) \]
\[ D_p = \begin{vmatrix} \alpha_1 & \beta_1 & P \\ \alpha_2 & \beta_2 & Q \\ \alpha_3 & \beta_3 & R \end{vmatrix} = \alpha_1(\beta_2 R - Q\beta_3) - \beta_1(\alpha_2 R - Q\alpha_3) + P(\alpha_2\beta_3 - \beta_2\alpha_3) \]
6. Rumus Akhir
Substitusikan kembali notasi asli:
\[ m = \frac{D_m}{D}, \qquad n = \frac{D_n}{D}, \qquad p = \frac{D_p}{D} \]
dengan \(D\), \(D_m\), \(D_n\), \(D_p\) seperti didefinisikan di atas, di mana:
\(P = \log\dfrac{v_2}{v_1},\ Q = \log\dfrac{v_3}{v_1},\ R = \log\dfrac{v_4}{v_1}\)
\(\alpha_i = \log\dfrac{[A]_{i+1}}{[A]_1},\ \beta_i = \log\dfrac{[B]_{i+1}}{[B]_1},\ \gamma_i = \log\dfrac{[C]_{i+1}}{[C]_1}\quad (i=1,2,3)\)
7. Hubungan dengan Kasus Dua Pereaksi
Rumus untuk dua pereaksi adalah kasus khusus ketika \(p = 0\) dan kolom \(\gamma\) dihilangkan, sistem menyusut kembali menjadi 2×2. Pola penurunannya identik, hanya dimensinya yang berbeda.
Demikian pula, bila data percobaan dirancang sehingga beberapa konsentrasi tidak berubah, sejumlah elemen log akan bernilai nol, dan determinan-determinan di atas akan menyederhanakan diri secara alami menjadi rumus yang lebih ringkas.
Contoh 1
Data percobaan reaksi A + B + C → produk:
| Perc. | [A] (mol/L) |
[B] (mol/L) |
[C] (mol/L) |
v (mol L−1 s−1) |
|---|---|---|---|---|
| 1 | 0,10 | 0,20 | 0,10 | 2,0 × 10−4 |
| 2 | 0,20 | 0,20 | 0,10 | 4,0 × 10−4 |
| 3 | 0,20 | 0,40 | 0,10 | 1,6 × 10−3 |
| 4 | 0,20 | 0,40 | 0,20 | 3,2 × 10−3 |
Penyelesaian
Langkah 1 — Hitung nilai-nilai logaritma
\[ P = \log\frac{4{,}0\times10^{-4}}{2{,}0\times10^{-4}} = \log 2 = 0{,}3010 \] \[ Q = \log\frac{1{,}6\times10^{-3}}{2{,}0\times10^{-4}} = \log 8 = 0{,}9031 \] \[ R = \log\frac{3{,}2\times10^{-3}}{2{,}0\times10^{-4}} = \log 16 = 1{,}2041 \]
\[ \alpha_1 = \log\frac{0{,}20}{0{,}10} = \log 2 = 0{,}3010 \quad \alpha_2 = \log\frac{0{,}20}{0{,}10} = \log 2 = 0{,}3010 \quad \alpha_3 = \log\frac{0{,}20}{0{,}10} = \log 2 = 0{,}3010 \] \[ \beta_1 = \log\frac{0{,}20}{0{,}20} = \log 1 = 0 \quad \beta_2 = \log\frac{0{,}40}{0{,}20} = \log 2 = 0{,}3010 \quad \beta_3 = \log\frac{0{,}40}{0{,}20} = \log 2 = 0{,}3010 \] \[ \gamma_1 = \log\frac{0{,}10}{0{,}10} = \log 1 = 0 \quad \gamma_2 = \log\frac{0{,}10}{0{,}10} = \log 1 = 0 \quad \gamma_3 = \log\frac{0{,}20}{0{,}10} = \log 2 = 0{,}3010 \]
Langkah 2 — Hitung tiga minor yang dipakai berulang
\[ M_{11} = \beta_2\gamma_3 - \gamma_2\beta_3 = (0{,}3010)(0{,}3010) - (0)(0{,}3010) = 0{,}09060 \] \[ M_{12} = \alpha_2\gamma_3 - \gamma_2\alpha_3 = (0{,}3010)(0{,}3010) - (0)(0{,}3010) = 0{,}09060 \] \[ M_{13} = \alpha_2\beta_3 - \beta_2\alpha_3 = (0{,}3010)(0{,}3010) - (0{,}3010)(0{,}3010) = 0 \]
Langkah 3 — Hitung \(D\)
\[ D = \alpha_1 M_{11} - \beta_1 M_{12} + \gamma_1 M_{13} = (0{,}3010)(0{,}09060) - (0)(0{,}09060) + (0)(0) = 0{,}02727 \]
Langkah 4 — Hitung \(D_m\)
\[ D_m = P\,M_{11} - \beta_1(Q\gamma_3 - \gamma_2 R) + \gamma_1(Q\beta_3 - \beta_2 R) \] \[ = (0{,}3010)(0{,}09060) - (0)(\ldots) + (0)(\ldots) = 0{,}02727 \] \[ m = \frac{D_m}{D} = \frac{0{,}02727}{0{,}02727} = \mathbf{1} \]
Langkah 5 — Hitung \(D_n\)
\[ D_n = \alpha_1(Q\gamma_3 - \gamma_2 R) - P\,M_{12} + \gamma_1(\alpha_2 R - Q\alpha_3) \] \[ = (0{,}3010)\bigl[(0{,}9031)(0{,}3010)-(0)(1{,}2041)\bigr] - (0{,}3010)(0{,}09060) + (0)(\ldots) \] \[ = (0{,}3010)(0{,}27183) - 0{,}02727 = 0{,}08182 - 0{,}02727 = 0{,}05455 \] \[ n = \frac{D_n}{D} = \frac{0{,}05455}{0{,}02727} = \mathbf{2} \]
Langkah 6 — Hitung \(D_p\)
\[ D_p = \alpha_1(\beta_2 R - Q\beta_3) - \beta_1(\alpha_2 R - Q\alpha_3) + P\,M_{13} \] \[ = (0{,}3010)\bigl[(0{,}3010)(1{,}2041)-(0{,}9031)(0{,}3010)\bigr] - (0)(\ldots) + (0{,}3010)(0) \] \[ = (0{,}3010)(0{,}36243 - 0{,}27183) = (0{,}3010)(0{,}09060) = 0{,}02727 \] \[ p = \frac{D_p}{D} = \frac{0{,}02727}{0{,}02727} = \mathbf{1} \]
Hasil
\[ m = 1,\quad n = 2,\quad p = 1 \]
\[ \text{Orde total} = m + n + p = 1 + 2 + 1 = \mathbf{4} \]
\[ k = \frac{v_1}{[A]_1^m[B]_1^n[C]_1^p} = \frac{2{,}0\times10^{-4}}{(0{,}10)^1(0{,}20)^2(0{,}10)^1} = \frac{2{,}0\times10^{-4}}{4{,}0\times10^{-4}} = 0{,}50 \] Cek dengan percobaan 4: \[ k = \frac{3{,}2\times10^{-3}}{(0{,}20)^1(0{,}40)^2(0{,}20)^1} = \frac{3{,}2\times10^{-3}}{6{,}4\times10^{-3}} = 0{,}50 \checkmark \]
Contoh 2 — Semua Konsentrasi Berbeda
Data percobaan reaksi A + B + C → produk:
| Perc. | [A] (mol/L) |
[B] (mol/L) |
[C] (mol/L) |
v (mol L−1 s−1) |
|---|---|---|---|---|
| 1 | 0,10 | 0,10 | 0,10 | 2,0 × 10−4 |
| 2 | 0,20 | 0,15 | 0,20 | 2,4 × 10−3 |
| 3 | 0,30 | 0,20 | 0,10 | 1,2 × 10−3 |
| 4 | 0,10 | 0,20 | 0,30 | 3,6 × 10−3 |
Penyelesaian
Semua konsentrasi berbeda di setiap percobaan. Rumus umum Cramer 3×3 harus digunakan.
Langkah 1 — Hitung nilai-nilai logaritma
\[ P = \log\frac{2{,}4\times10^{-3}}{2{,}0\times10^{-4}} = \log 12 = 1{,}0792 \] \[ Q = \log\frac{1{,}2\times10^{-3}}{2{,}0\times10^{-4}} = \log 6 = 0{,}7782 \] \[ R = \log\frac{3{,}6\times10^{-3}}{2{,}0\times10^{-4}} = \log 18 = 1{,}2553 \]
\[ \alpha_1 = \log\frac{0{,}20}{0{,}10} = \log 2 = 0{,}3010 \quad \alpha_2 = \log\frac{0{,}30}{0{,}10} = \log 3 = 0{,}4771 \quad \alpha_3 = \log\frac{0{,}10}{0{,}10} = \log 1 = 0 \] \[ \beta_1 = \log\frac{0{,}15}{0{,}10} = \log 1{,}5 = 0{,}1761 \quad \beta_2 = \log\frac{0{,}20}{0{,}10} = \log 2 = 0{,}3010 \quad \beta_3 = \log\frac{0{,}20}{0{,}10} = \log 2 = 0{,}3010 \] \[ \gamma_1 = \log\frac{0{,}20}{0{,}10} = \log 2 = 0{,}3010 \quad \gamma_2 = \log\frac{0{,}10}{0{,}10} = \log 1 = 0 \quad \gamma_3 = \log\frac{0{,}30}{0{,}10} = \log 3 = 0{,}4771 \]
Langkah 2 — Hitung tiga minor
\[ M_{11} = \beta_2\gamma_3 - \gamma_2\beta_3 = (0{,}3010)(0{,}4771) - (0)(0{,}3010) = 0{,}14363 \] \[ M_{12} = \alpha_2\gamma_3 - \gamma_2\alpha_3 = (0{,}4771)(0{,}4771) - (0)(0) = 0{,}22763 \] \[ M_{13} = \alpha_2\beta_3 - \beta_2\alpha_3 = (0{,}4771)(0{,}3010) - (0{,}3010)(0) = 0{,}14363 \]
Langkah 3 — Hitung \(D\)
\[ D = \alpha_1 M_{11} - \beta_1 M_{12} + \gamma_1 M_{13} \] \[ = (0{,}3010)(0{,}14363) - (0{,}1761)(0{,}22763) + (0{,}3010)(0{,}14363) \] \[ = 0{,}04323 - 0{,}04009 + 0{,}04323 = 0{,}04637 \]
Langkah 4 — Hitung \(D_m\)
\[ D_m = P\,M_{11} - \beta_1(Q\gamma_3 - \gamma_2 R) + \gamma_1(Q\beta_3 - \beta_2 R) \]
Hitung suku-suku dalam kurung:
\[ Q\gamma_3 - \gamma_2 R = (0{,}7782)(0{,}4771) - (0)(1{,}2553) = 0{,}37131 \] \[ Q\beta_3 - \beta_2 R = (0{,}7782)(0{,}3010) - (0{,}3010)(1{,}2553) = 0{,}23424 - 0{,}37784 = -0{,}14360 \]Substitusikan:
\[ D_m = (1{,}0792)(0{,}14363) - (0{,}1761)(0{,}37131) + (0{,}3010)(-0{,}14360) \] \[ = 0{,}15497 - 0{,}06539 - 0{,}04322 = 0{,}04636 \] \[ m = \frac{D_m}{D} = \frac{0{,}04636}{0{,}04637} \approx \mathbf{1} \]Langkah 5 — Hitung \(D_n\)
\[ D_n = \alpha_1(Q\gamma_3 - \gamma_2 R) - P\,M_{12} + \gamma_1(\alpha_2 R - Q\alpha_3) \]
Hitung suku baru:
\[ \alpha_2 R - Q\alpha_3 = (0{,}4771)(1{,}2553) - (0{,}7782)(0) = 0{,}59892 \]Substitusikan:
\[ D_n = (0{,}3010)(0{,}37131) - (1{,}0792)(0{,}22763) + (0{,}3010)(0{,}59892) \] \[ = 0{,}11177 - 0{,}24566 + 0{,}18028 = 0{,}04639 \] \[ n = \frac{D_n}{D} = \frac{0{,}04639}{0{,}04637} \approx \mathbf{1} \]Langkah 6 — Hitung \(D_p\)
\[ D_p = \alpha_1(\beta_2 R - Q\beta_3) - \beta_1(\alpha_2 R - Q\alpha_3) + P\,M_{13} \]
Hitung suku baru:
\[ \beta_2 R - Q\beta_3 = (0{,}3010)(1{,}2553) - (0{,}7782)(0{,}3010) = 0{,}37784 - 0{,}23424 = 0{,}14360 \]Substitusikan:
\[ D_p = (0{,}3010)(0{,}14360) - (0{,}1761)(0{,}59892) + (1{,}0792)(0{,}14363) \] \[ = 0{,}04322 - 0{,}10547 + 0{,}15498 = 0{,}09273 \] \[ p = \frac{D_p}{D} = \frac{0{,}09273}{0{,}04637} \approx \mathbf{2} \]Hasil
\[ m = 1,\quad n = 1,\quad p = 2 \]
\[ \text{Orde total} = m + n + p = 1 + 1 + 2 = \mathbf{4} \]
\[ k_1 = \frac{2{,}0\times10^{-4}}{(0{,}10)^1(0{,}10)^1(0{,}10)^2} = \frac{2{,}0\times10^{-4}}{1{,}0\times10^{-4}} = 2{,}0 \] \[ k_2 = \frac{2{,}4\times10^{-3}}{(0{,}20)^1(0{,}15)^1(0{,}20)^2} = \frac{2{,}4\times10^{-3}}{1{,}2\times10^{-3}} = 2{,}0 \] \[ k_3 = \frac{1{,}2\times10^{-3}}{(0{,}30)^1(0{,}20)^1(0{,}10)^2} = \frac{1{,}2\times10^{-3}}{6{,}0\times10^{-4}} = 2{,}0 \] \[ k_4 = \frac{3{,}6\times10^{-3}}{(0{,}10)^1(0{,}20)^1(0{,}30)^2} = \frac{3{,}6\times10^{-3}}{1{,}8\times10^{-3}} = 2{,}0 \]
Keempat percobaan menghasilkan nilai \(k\) yang identik — membuktikan bahwa hasil \(m\), \(n\), dan \(p\) yang diperoleh secara matematis sempurna konsisten dengan seluruh data.
Ringkasan Pola Umum
Penurunan untuk dua dan tiga pereaksi mengikuti pola yang sama persis:
| Jumlah pereaksi |
Bilangan tidak diketahui |
Percobaan minimum |
Dimensi sistem |
|---|---|---|---|
| 2 | \(m,\ n\) | 3 | 2×2 |
| 3 | \(m,\ n,\ p\) | 4 | 3×3 |
| \(r\) | \(r\) orde | \(r+1\) | \(r\times r\) |
Langkahnya selalu sama: ambil logaritma → tulis untuk \(r+1\) percobaan → kurangkan percobaan 1 dari semua yang lain untuk eliminasi \(\log k\) → susun sistem \(r \times r\) → selesaikan dengan Aturan Cramer.
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