Química, perguntado por anonimosilva351, 10 meses atrás

Sejam dadas as equações termoquímicas, todas a 25 ºC e 1 atm:

I- H2(g)+ ½ O2(g) →H2O(l) ∆H = -68,3 Kcal/mol
II- 2Fe(s)+ 3/2 O2(g)→Fe2O3(s) ∆H = -196,5 Kcal/mol
III- 2Al(s)+ 3/2 O2(g)→Al2O3(s) ∆H = -399,1 Kcal/mol
IV - C(grafite)+ O2(g)→ CO2(g) ∆H = -94,0 Kcal/mol
V- CH4(g) + O2(g) → CO2(g)+ H2O(l) ∆H = -17,9 Kcal/mol

Exclusivamente sob o ponto de vista energético, das reações acima, a que você escolheria como fonte de energia é:
b) I d) II p) III q) IV t) V

Soluções para a tarefa

Respondido por Thoth
4

Resposta:

Alternativa  p) III

Explicação:

III- 2Al(s)+ 3/2 O2(g)→Al2O3(s) ∆H = -399,1 Kcal/mol, porque é a reação que libera maior quantidade de calor (399,1 Kcal/mol).


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