From the following data at 25C- Calculate the standard enthalpy of formation of FeOs and of Fe 2 O 3 s -Reaction - r H KJ-mole1 Fe 2 O 3 s -3C graphite - 2Fe s -CO g 4922 FeO s -C graphite - Fe s -CO g 1553 C graphite - O 2 g - C O 2 g -3934 CO g - 1 - 2 O 2 g - C O 2 g -282

By Hess Law,For finding the standard enthalpy of formation of FeO(s) Fe(s)+CO(g)→ C(graphite)+Fe O(s) #160; #160; #160; #160; #160; ...

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Heat of Formation

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Question

From the following data at 25C, Calculate the standard enthalpy of formation of FeO(s) and of ${F e}_{2} O_{3} (s)$ .
Reaction $Δ_{r} H (K J / m o l e)$
(1) ${F e}_{2} O_{3} (s) + 3 C (g r a p h i t e) \to 2 F e (s) + C O (g)$ 492
(2) $F e O (s) + C (g r a p h i t e) \to F e (s) + C O (g)$ 155
(3) $C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g)$ -393
(4) $C O (g) + 1 / 2 O_{2} (g) \to C O_{2} (g)$ -282

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25^{o} C

{F e}_{2} O_{3} (s) + 3 C_{g r a p h i t e} ⟶ 2 F e (s) + 3 C O (g); Δ H^{o} = 492 k J / m o l

F e O (s) + C_{g r a p h i t e} ⟶ F e (s) + C O (g); Δ H^{o} = 155.8 k J / m o l

C_{g r a p h i t e} + O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 393.51 k J / m o l

C O (g) + \frac{1}{2} O_{2} (g) ⟶ C O_{2} (g); Δ H^{o} = - 282.98 k J / m o l

F e O (s)

{F e}_{2} O_{3} (s)

25^{\circ} C

C H_{3} O H (I)

C H_{3} O H (I) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l); △_{r} H^{θ} = - 726 k J m p l^{- 1}

C_{(g r a p h i t e)} + O_{2} (g) \to C O_{2} (g); △_{c} H^{θ} = - 393 k J m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); △_{f} H^{θ} = - 286 k J m o l^{- 1}

F e O (s) + C (g r a p h i t e) ⟶ F e (s) + C O (g)

C H_{3} O H (1) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (I); Δ, H^{⊖} = - 726 k J m o l^{- 1}

C (g r a p h i t e) + O_{2} (g) \to C O_{2} (g); Δ_{c} H^{⊖} = - 393 k J m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (I); Δ_{f} H^{⊖} = - 286 k J m o l^{- 1}

{C H}_{3} O H (l)

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