Ejercicio 6
Planteo:
A + B → k C A + B \xrightarrow{k} C A + B k C k k k Ley de acción de masa : La tasa de cambio de una reacción química elemental es proporcional al producto de las concentraciones de los reactantes.
↪ \hookrightarrow ↪
A + B → k C ⟹ tasa cambio = k A B ⟹ { A ˙ = − k A B B ˙ = − k A B C ˙ = k A B A + B \xrightarrow{k} C \implies \text{tasa cambio} = kAB \implies \begin{cases} \dot{A} = -kAB \\ \dot{B} = -kAB \\ \dot{C} = kAB \end{cases} A + B k C ⟹ tasa cambio = k A B ⟹ ⎩ ⎨ ⎧ A ˙ = − k A B B ˙ = − k A B C ˙ = k A B (Nota lateral en imagen: A 0 − B 0 = c t e ⟹ A − B = c t e A_0 - B_0 = cte \implies A - B = cte A 0 − B 0 = c t e ⟹ A − B = c t e
Resolución:
↪ A − B = N ⟹ A = N + B ⟹ B ˙ = − k ( N + B ) B ⟹ B ˙ = − k N B − k B 2 \hookrightarrow A - B = N \implies A = N + B \implies \dot{B} = -k(N+B)B \implies \dot{B} = -kNB - kB^2 ↪ A − B = N ⟹ A = N + B ⟹ B ˙ = − k ( N + B ) B ⟹ B ˙ = − k N B − k B 2 ⟹ d B − k N B − k B 2 = d t ⟹ ∫ d B − k N B − k B 2 = ∫ d t ⟹ ∫ d B k B ( − N − B ) = ∫ d t \implies \frac{dB}{-kNB - kB^2} = dt \implies \int \frac{dB}{-kNB - kB^2} = \int dt \implies \int \frac{dB}{kB(-N-B)} = \int dt ⟹ − k N B − k B 2 d B = d t ⟹ ∫ − k N B − k B 2 d B = ∫ d t ⟹ ∫ k B ( − N − B ) d B = ∫ d t ⟹ ∫ d B B ( N + B ) = − k ∫ d t \implies \int \frac{dB}{B(N+B)} = -k \int dt ⟹ ∫ B ( N + B ) d B = − k ∫ d t Resolviendo por fracciones simples:
∣ 1 B ( N + B ) = α B + γ N + B ⟹ 1 = ( N + B ) α + B γ \left| \frac{1}{B(N+B)} = \frac{\alpha}{B} + \frac{\gamma}{N+B} \implies 1 = (N+B)\alpha + B\gamma \right. B ( N + B ) 1 = B α + N + B γ ⟹ 1 = ( N + B ) α + B γ B = 0 ⟹ 1 = N α ⟹ α = 1 / N B = 0 \implies 1 = N\alpha \implies \alpha = 1/N B = 0 ⟹ 1 = N α ⟹ α = 1/ N B = − N ⟹ 1 = − N γ ⟹ γ = − 1 / N B = -N \implies 1 = -N\gamma \implies \gamma = -1/N B = − N ⟹ 1 = − N γ ⟹ γ = − 1/ N Sustituyendo en la integral:
⟹ ∫ 1 / N B d B − ∫ 1 / N N + B d B = − k t + c \implies \int \frac{1/N}{B} dB - \int \frac{1/N}{N+B} dB = -kt + c ⟹ ∫ B 1/ N d B − ∫ N + B 1/ N d B = − k t + c ln ( ∣ B ∣ ) N − ln ( ∣ N + B ∣ ) N = − k t + c ~ \frac{\ln(|B|)}{N} - \frac{\ln(|N+B|)}{N} = -kt + \tilde{c} N ln ( ∣ B ∣ ) − N ln ( ∣ N + B ∣ ) = − k t + c ~ ⟹ 1 N ln ∣ B N + B ∣ = − k t + c ~ ⟹ ln ∣ B N + B ∣ = − N k t + c ^ \implies \frac{1}{N} \ln \left| \frac{B}{N+B} \right| = -kt + \tilde{c} \implies \ln \left| \frac{B}{N+B} \right| = -Nkt + \hat{c} ⟹ N 1 ln N + B B = − k t + c ~ ⟹ ln N + B B = − N k t + c ^ ⟹ B N + B = C e − N k t ⟹ B = N C e − N k t + B C e − N k t \implies \frac{B}{N+B} = C e^{-Nkt} \implies B = N C e^{-Nkt} + B C e^{-Nkt} ⟹ N + B B = C e − N k t ⟹ B = N C e − N k t + B C e − N k t ⟹ B ( 1 − C e − N k t ) = N C e − N k t ⟹ B ( t ) = N C e − N k t 1 − C e − N k t \implies B(1 - C e^{-Nkt}) = N C e^{-Nkt} \implies B(t) = \frac{N C e^{-Nkt}}{1 - C e^{-Nkt}} ⟹ B ( 1 − C e − N k t ) = N C e − N k t ⟹ B ( t ) = 1 − C e − N k t N C e − N k t ⟹ A ( t ) = N + B ( t ) = N ( 1 + C e − N k t 1 − C e − N k t ) = N 1 − C e − N k t \implies A(t) = N + B(t) = N \left( 1 + \frac{C e^{-Nkt}}{1 - C e^{-Nkt}} \right) = \frac{N}{1 - C e^{-Nkt}} ⟹ A ( t ) = N + B ( t ) = N ( 1 + 1 − C e − N k t C e − N k t ) = 1 − C e − N k t N