Question

Calcule o ΔH da reação abaixo:

C₃H₈ + 5O₂ --> 3CO₂ = 4H₂O

a partir das reações intermediárias:
a) 3C + 4H₂ --> C₃H₈ ΔH = -103,7 kJ
b) C + O₂ --> CO₂ ΔH = -392,9 kJ
c) H₂ + 1/2 O₂ --> H₂O ΔH = -241,6 kJ

Answer 1

$\boxed{\begin{array}{l}\sf l: inverta~a~1^a~reac_{\!\!,}\tilde ao\\\sf e~troque~o~sinal~do~\Delta\,h:\\\sf C_3H_8\longrightarrow3C+4H_2~~\Delta\,H=+103,7\,KJ\\\sf multiplique~a~2^a~reac_{\!\!,}\tilde ao~por~3,\\\sf e~multiplique~o~\Delta\,H~por~3:\\\sf 3C+3O_2\longrightarrow 3CO_2~~\Delta\,H=+178,7~KJ\\\sf primeiro~escreva~a~3^a~reac_{\!\!,}\tilde ao~com~os~menores\\\sf coeficientes~ inteiros:\\\sf 2H_2+O_2\longrightarrow 2H_2O\end{array}}$

$\boxed{\begin{array}{l}\sf multiplique~a~reac_{\!\!,}\tilde ao~anterior~por~2\\\sf e~o~\Delta\,H~da~reac_{\!\!,}\tilde ao~tamb\acute em~por~2:\\\sf 4H_2+2O_2\longrightarrow 4H_2O~~\Delta\,H=-966,4~KJ\\\underline{\rm Vamos~adicionar~as~equac_{\!\!,}\tilde oes~membro}\\\underline{\rm a~membro~e~calcular~o~\Delta\,H~pedido:}\\+\underline{\begin{cases}\sf C_3H_8\longrightarrow \diagdown\!\!\!\!\!\!3C+ \diagdown\!\!\!\!\!\!4H_2~~\Delta\,H=103,7~KJ\\\sf\diagdown\!\!\!\!\!\!3C+3O_2\longrightarrow 3CO_2~~\Delta\,H=+178,7~KJ\\\sf\diagdown\!\!\!\!\!\!4H_2+2O_2\longrightarrow 4H_2O~~\Delta\,H=-966,4~KJ\end{cases}}\\\sf C_3H_8+5O_2\longrightarrow 3CO_2+4H_2O\\\sf\Delta\,H=103,7+178,7-966,4\\\huge\boxed{\boxed{\boxed{\boxed{\sf\Delta\,H=-684~KJ}}}} \end{array}}$