Para a reação em fase gasosa 2 \mathrm{HI} \rightarrow \mathrm{H}_{2}+\mathrm{I}_{2}, os valores de k são 1,2 \times 10^{-3} \mathrm{e} 3,0 \times 10^{-5} \mathrm{dm}^{3} \mathrm{~mol}^{-1} \mathrm{s}^{-1} a 700 e 629 K , respectivamente. Calcule \mathrm{E}_{\mathrm{a}} e \mathrm{A} .\left(\right. Resp.: \mathrm{E}_{\mathrm{a}}=190,6 \mathrm{KJ} / \mathrm{mol} e \mathrm{A}=1,86 \times 10^{11} \mathrm{dm}^{3} \mathrm{mol}^{-1} \mathrm{~s}^{-1} )

Passo 1

O enunciado dá a reação gasosa 2\,\mathrm{HI}\to\mathrm{H_2}+\mathrm{I_2} com constantes de velocidade \,k_1=1,2\times10^{-3}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}} a T_1=700\,\mathrm{K} e \,k_2=3,0\times10^{-5}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}} a T_2=629\,\mathrm{K}.
Queremos E_a e A usando a lei de Arrhenius
k = A \exp\!\bigl(-\tfrac{E_a}{R\,T}\bigr).

Passo 2

Tomamos a razão de ks e linearizamos para isolar E_a:
\ln\frac{k_2}{k_1} =-\frac{E_a}{R}\Bigl(\frac{1}{T_2}-\frac{1}{T_1}\Bigr).
Com R=8,314\,\mathrm{J\,mol^{-1}\,K^{-1}},
\ln\frac{k_2}{k_1} =\ln\frac{3,0\times10^{-5}}{1,2\times10^{-3}} =\ln(0,025)=-3,689,
\frac{1}{T_2}-\frac{1}{T_1} =\frac{1}{629}-\frac{1}{700} =1,589\times10^{-3}-1,429\times10^{-3} =1,604\times10^{-4}\,\mathrm{K^{-1}}.
Logo,
E_a =-R\;\frac{\ln(k_2/k_1)}{\tfrac{1}{T_2}-\tfrac{1}{T_1}} =-8,314\cdot\frac{-3,689}{1,604\times10^{-4}} \approx1,906\times10^{5}\,\mathrm{J/mol} =190,6\,\mathrm{kJ/mol}.

Passo 3

Agora calculamos o fator pré-exponencial A a partir de um dos k, digamos k_1:
A =k_1\exp\!\Bigl(\frac{E_a}{R\,T_1}\Bigr).
Substituindo E_a=1,906\times10^{5}\,\mathrm{J/mol}, T_1=700\,\mathrm{K} e k_1=1,2\times10^{-3},
A =1,2\times10^{-3}\exp\!\Bigl(\frac{1,906\times10^{5}}{8,314\times700}\Bigr) \approx1,86\times10^{11}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}.

Resposta

E_a=190,6\,\mathrm{kJ/mol} A=1,86\times10^{11}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}

Exercícios relacionados