Calculate- the - H at 85oC for the reaction- Fe 2 O 3 s-3 H 2 - 2Fes-3 H 2 OlThe data are- - H 298 o -33-29 kJ-mol andSubstance Fe2O3s Fes H2Ol H2g C P o J-Kmol 103-8 25-1 75-3 28-8

Fe _ 2 O _ 3 (s)+3 H _ 2 ⟶ 2Fe(s)+3 H _ 2 O(l) Δ C _ P =( C _ P ) #160;_ P ( C _ P ) #160;_ R =(2× 25.1+3± 75.3) (103.8+3× 28.8)=85.9J ...

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Calculate, the $Δ H$ at $85^{o} C$ for the reaction:
${F e}_{2} O_{3} (s) + 3 H_{2} ⟶ 2 F e (s) + 3 H_{2} O (l)$
The data are: $Δ H_{298}^{o} = - 33.29 k J / m o l$ and
Substance ${F e}_{2} O_{3} (s)$ $F e (s)$ $H_{2} O (l)$ $H_{2} (g)$
$C_{P}^{o} (J / K m o l)$ $103.8$ $25.1$ $75.3$ $28.8$

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Δ H

358 K

F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 3 H_{2} O (l)

Δ H_{298} = - 33.29 k J m o l^{- 1}

C_{P} f o r F e_{2} O_{3} (s), F e (s), H_{2} O (l)

H_{2} (g)

103.8, 25.1, 75.3, a n d 28.8 J / K m o l

H_{2} O (l)

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

- 286

k J

{m o l}^{- 1}

- 824

k J

{m o l}^{- 1}

{F e}_{2} O_{3} (s) + 3 H_{2} (g) ⟶ 3 H_{2} O (l) + 2 F e (s)

{F e}_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 3 H_{2} O (l)

Δ H_{300}^{o} = - 26.72 k J

Δ H^{o}

C_{p} [{F e}_{2} O_{3}] = 105, C_{p} [F e (s)] = 25, C_{p} [H_{2} O (l)] = 75, C_{p} [H_{2} (g)] = 30 (all are in J/mol)

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

2 F e (s) + \frac{3}{2} O_{2} (g) ⟶ {F e}_{2} O_{3} (s); Δ H = - 193.4 k J . . . (i)

M g (s) + \frac{1}{2} O_{2} (g) ⟶ M g O (s); Δ H = - 140.2 k J . . . . (i i)

Δ H

3 M g + {F e}_{2} O_{3} ⟶ 3 M g O + 2 F e

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Substance	${F e}_{2} O_{3} (s)$	$F e (s)$	$H_{2} O (l)$	$H_{2} (g)$
$C_{P}^{o} (J / K m o l)$	$103.8$	$25.1$	$75.3$	$28.8$