Calculate the heat of formation of CH3COOHl at 25-C from the following data- CH3COOHl-2O2g - 2CO2 -2H2Ol- - H-208-34 kcal Cs -O2g - CO2g- H-94-05 kcal H2g - 12O2- H2Ol- - H-68-32 kcal

(1) CH_3COOH(l)+2O_2(g) → 2CO_2 +2H_2O(l); Δ H_1= 208.34 kcal(2) C(s) +O_2(g) → #160; CO_2(g);Δ H_2= 94.05 kcal(3) H_2(g) + 1/2O_2→ H_2O(l); ...

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Third Law of Thermodynamics

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Question

Calculate the heat of formation of $C H_{3} C O O H (l)$ at $25^{\circ} C$ from the following data;
$C H_{3} C O O H (l) + 2 O_{2} (g) \to 2 C O_{2} + 2 H_{2} O (l); Δ H = - 208.34$ kcal
$C (s) + O_{2} (g) \to C O_{2} (g); Δ H = - 94.05$ kcal
$H_{2} (g) + \frac{1}{2} O_{2} \to H_{2} O (l); Δ H = - 68.32$ kcal

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{C H}_{3} C O O H (l) + 2 O_{2} (g) \to 2 {C O}_{2} + 2 H_{2} O (l) . . . . (i) (Δ H = - 2079 k c a l)

C (s) + O_{2} (g) \to {C O}_{2} (g); (Δ H = - 94.48 k c a l) . . . (i i)

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); (Δ H = - 68.4 k c a l) . . . . (i i i)

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 H}_{3} O H (l)

{C H}_{3} O H (l) + \frac{3}{2} O_{2} (g) ⟶ C O_{2} (g) + 2 H_{2} O (l); Δ H = - 726 k J

C (s) + O_{2} (g) ⟶ C O_{2} (g); Δ H = - 394 k J

H_{2} (g) + \frac{1}{2} O_{2} (g) ⟶ H_{2} O (l); Δ H = - 286 k J

C_{2} H_{2}

C_{2} H_{4}

H_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ H_{2} O (l); Δ H = - 68.32 k c a l

C_{2} H_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ 2 C O_{2} (g) + H_{2} O (l); Δ H = - 310.61 k c a l

C_{2} H_{4} (g) + \frac{3}{2} O_{2} (g) ⇌ 2 C O_{2} (g) + 2 H_{2} O (l); Δ H = - 337.32 k c a l

C H_{3} O H (l)

C H_{3} O H (l) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l); △_{r} H^{⊖} = - 726 k J

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

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), Δ H = - 286 k J

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