Autor: química e você! | Publicado: 2012-08-10T12:53:00-07:00

Utilizando calor de formação


Toda substância apresenta um valor tabelado de entalpia de formação,de maneira que é possível calcular o&nbsp;∆H de uma reação qualquer,utilizando esses valores,sabendo que : &nbsp;


&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;
&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;



Os coeficientes estequiométricos de cada componente devem ser multiplicados pelos respectivos valores de calor de formação.





&nbsp;A entalpia de formação de uma substância simples no estado padrão&nbsp;é sempre zero.&nbsp;





Utilizando energia de ligação




Toda reação química ocorre com a ruptura de ligações e estabelecimento de novas ligações.Nesse contexto é importante saber:






Lado dos reagentes: &nbsp;*ruptura de ligações&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;*absorção de energia (ENDOTÉRMICA)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;*sinal das energias será positivo (+)
Lado dos produtos: *estabelecimento de ligações&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;*liberação de energia (EXOTÉRMICA)&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;*sinal das energias será negativo(-)





Exemplo:&nbsp;




Cl2 (g) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;2 Cl (g) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;DH &nbsp;= &nbsp;58 Kcal

&nbsp;N2 (g) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; 2 N (g) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; DH &nbsp;= &nbsp;226 Kcal

&nbsp;HCl (aq) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; H (g) &nbsp; + &nbsp; Cl (g) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; DH &nbsp;= &nbsp;103 Kcal

Calcule o DH da reação: &nbsp; &nbsp; &nbsp;C2H4 (g) &nbsp; + &nbsp; H2 (g) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; C2H6 (g) , dadas as energias de ligação em kcal: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;

&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;C = C ... 146,8 &nbsp;kcal/mol &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;C – C ... &nbsp; 83,2 &nbsp;kcal/mol

&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;C – H ... &nbsp; 98,8 &nbsp;kcal/mol &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;H – H ... 104,2 &nbsp;kcal/mol

&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; H &nbsp; H

&nbsp; H &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;H

&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;C = C &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; + &nbsp; &nbsp; H &nbsp;– &nbsp;H &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;H – C – C – H&nbsp;

&nbsp; H &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;H

&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;H &nbsp; H

&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;

Total de ligações &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;Total de ligações

&nbsp; &nbsp; quebradas &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;formadas

DH &nbsp;= &nbsp;4 . 98,8 &nbsp; + &nbsp;1 . 146,8 &nbsp;+ &nbsp;1 . 104,2 &nbsp;+ &nbsp;6 . (– 98,8) &nbsp;+ &nbsp;1 . (– 83,2)

DH &nbsp;= &nbsp;– 29,8 kcal

4 . C – H

1 . C = C

1 . H – H &nbsp;

6 . C – H

1 . C – C




DH &nbsp;= &nbsp;– 29,8 kcal




Postado por: Fabiane Coelho