Questões do ENEM
Autor: Quipibid | Publicado: 2017-09-09T13:32:00-03:00




Bolsista: Karla Matias



Olá queridos leitores, hoje vamos
resolver uma questão de química do ENEM/2016, com a intenção de tirar as
dúvidas quanto a resolução das questões.

Questão N° 50 Caderno AZUL. 

O benzeno, um importante solvente para a
indústria química, é obtido industrialmente pela destilação do petróleo.
Contudo, também pode ser sintetizado pela trimerização do acetileno catalisada
por ferro metálico sob altas temperaturas, conforme a equação química:



&nbsp;3 C2
H2 (g) à&nbsp; C6 H6 (l)



A energia envolvida nesse processo pode
ser calculada indiretamente pela variação de entalpia das reações de combustão
das substâncias participantes, nas mesmas condições experimentais:

&nbsp;I. C2
H2 (g) 5/2 O2 (g) à 2 CO2 (g)&nbsp; H2O (l)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Hc ° =&nbsp; - 310 kcal/mol 

II. C6 H6 (l)&nbsp; 15/2 O2 (g) à 6 CO2 (g)&nbsp; 3 H2 O (l)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Hc ° =&nbsp; - 780 kcal/mol 

A variação de entalpia do processo de
trimerização, em kcal, para a formação de um mol de benzeno é mais próxima de:

<!--[if !supportLists]-->a)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<!--[endif]-->– 1090

<!--[if !supportLists]-->b)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<!--[endif]-->– 150 

<!--[if !supportLists]-->c)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<!--[endif]-->– 50 

<!--[if !supportLists]-->d)&nbsp;&nbsp;&nbsp;&nbsp;
<!--[endif]-->+ 157 


<!--[if !supportLists]-->e)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<!--[endif]-->+ 470



Antes de resolvê-la, vamos relembrar de
alguns conteúdos importantes de termoquímica que envolve a resolução desta
questão.&nbsp;



Lei de Hess



“A variação de entalpia envolvida em uma reação
química, sob certas condições experimentais, depende exclusivamente da entalpia
inicial e final dos produtos, seja a reação executada diretamente em uma única
etapa ou indiretamente em uma série de etapas. ”

A principal utilidade
da Lei de Hess é que ela nos permite determinar a variação de entalpia
envolvida em reações diretas por meio da combinação de várias equações
químicas, cujos valores das variações de entalpia são conhecidos. Depois basta
somar esses valores para determinar o calor da reação direta desejada.

A lei de Hess, permite
alterar as equações químicas, como se fossem matemáticas, podendo somar,
subtrair, multiplicar ou dividir, o que for necessário (dentro dos padrões
matemáticos) para conseguir chegar na equação padrão. 



Observe a resolução da
questão. 



3 C2
H2 (g) à&nbsp; C6 H6 (l)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Equação global



I.
Equação

I. &nbsp;&nbsp;&nbsp;C2 H2 (g) 5/2 O2
(g) à 2 CO2
(g)&nbsp; H2O (l)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Hc ° =&nbsp; - 310 kcal/mol



Observe, que o C2 H2, está no
lado dos reagentes, bem como na equação global, por isso vamos manter a
equação. 

Observe, que na equação global temos 3 mols de C2 H2, sendo
assim iremos multiplicar esta I equação por 3.



I.&nbsp;&nbsp;&nbsp; 3 C2
H2 (g) 15/2 O2 (g)
à 6 CO2 (g)&nbsp; 3 H2O
(l)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Hc° = -
930 kcal/mol

II.
Equação

II. &nbsp;C6 H6 (l)&nbsp; 15/2 O2 (g) à 6 CO2
(g)&nbsp; 3 H2 O (l)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Hc ° =&nbsp; - 780 kcal/mol



Observe que nesta II equação temos o C6 H6,
porém, ele está no lado dos reagentes, e conforme a equação global ele
deve estra no lado dos produtos. Por isso vamos inverter
a equação, lembrando que quando fazemos isso, inverte
também o sinal da variação de entalpia (Hc°).



II. 6 CO2
(g)&nbsp; 3 H2 O (l)
à C6 H6
(l)&nbsp; 15/2 O2 (g)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Hc° = + 780
kcal/mol









































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<![endif]--><!--[if !vml]--><!--[endif]-->Agora que temos as equações na devida ordem que
precisamos, vamos eliminar aquilo que não estar presente na equação global.







Então galerinha, espero tê-los ajudado a
tirar quaisquer dúvidas, fique ligado que na próxima matéria teremos mais
resoluções das questões do ENEM/2016. 



ATÉ A PROXIMA!!!