Given: H_2 + 1/2 O_2 rightarrow H_2O : Delta | vedclass

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(A) The formation reaction of $C_2H_5OH$ is: $2C(s) + 3H_2(g) + 1/2 O_2(g) \rightarrow C_2H_5OH(l)$.Given equations:$(1) \ H_2 + 1/2 O_2 \rightarrow H

Vedclass · Accepted Answer

$-66.2$

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(A) The formation reaction of $C_2H_5OH$ is: $2C(s) + 3H_2(g) + 1/2 O_2(g) ightarrow C_2H_5OH(l)$.Given equations:$(1) \ H_2 + 1/2 O_2 ightarrow H_2O, \Delta H_1 = -68.4 \ 	ext{kcal}$$(2) \ C + O_2 ightarrow CO_2, \Delta H_2 = -94.0 \ 	ext{kcal}$$(3) \ C_2H_5OH + 3O_2 ightarrow 2CO_2 + 3H_2O, \Delta H_3 = -327.0 \ 	ext{kcal}$To get the formation reaction,perform: $2 	imes (2) + 3 	imes (1) - (3)$:$\Delta H_f = 2(-94.0) + 3(-68.4) - (-327.0)$$\Delta H_f = -188.0 - 205.2 + 327.0$$\Delta H_f = -393.2 + 327.0 = -66.2 \ 	ext{kcal}$.

Given: $H_2 + 1/2 O_2 \rightarrow H_2O : \Delta H = -68.4 \ \text{kcal}$,$C + O_2 \rightarrow CO_2 : \Delta H = -94.0 \ \text{kcal}$,and $C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O : \Delta H = -327.0 \ \text{kcal}$. Calculate the heat of formation of $C_2H_5OH$ in $\text{kcal}$.

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