Calculate the heat of formation of acetic acid from the following data-CH3COOHl - 2O2g - 2CO2g - 2H2Ol- H - -207-9 K calCs - O2g - CO2g H - -94-48 K calH2g - 1-2 O2g - H2Ol H - -68-4 K cal

The required equation is 2C(s) + 2H_2(g) + O_2(g) → CH_3 COOH(l) H = ? CH_3COOH(l) + 2O_2(g) → 2CO_2(g) + 2H_2O(l) H = 207.9 K cal ...

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Heat of Formation

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Question

Calculate the heat of formation of acetic acid from the following data.

$C H_{3} C O O H (l) + 2 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l), △ H = - 207.9 K c a l$
$C (s) + O_{2} (g) \to C O 2 (g) △ H = - 94.48 K c a l$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) △ H = - 68.4 K c a l$

117.86 K cal

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116.86 K cal

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-116.86 K cal

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C (s) + O_{2} (g) ⟶ C O_{2} (g)

Δ H = - 393.7 k J

H_{2} (g) + 1 / 2 O_{2} (g) ⟶ H_{2} O (l)

Δ H = - 285.8 k J

C H_{3} C O O H (l) + 2 O_{2} (g) ⟶ 2 C O_{2} (g) + 2 H_{2} O (l)

Δ H = - 873.2 k J

Heat of formation of $C H_{4}$ is:

If given:

$C (s) + O_{2} (g) \to C O_{2} (g) △ H = - 394 k J$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) △ H = - 284 k J$

$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l) △ H = - 892 k J$

C (S) + O_{2} (g) \to C O_{2} (g)

Δ H = - 94.0 K c a l

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)

Δ H = - 68.0 K c a l

C H_{3} C O O H (l) + 2 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l)

Δ H = - 210.0 K c a l

C (s) + O_{2} (g) \to C O_{2} (g) Δ H = - 94.0 k c a l

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l) Δ H = - 68.0 k c a l

C H_{3} C O O H (l) + 2 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l) Δ H = - 210.0 k c a l

C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g); Δ H = - 393.5 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 286.2 k J

C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l); Δ H = - 1410.8 k J

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