Calculate- the - Go at 85 oC for the reaction-Fe2O3s-3H2g- 2Fes-3H2O The data are- - Ho298-35 kJ-mol- So358-0-095 kJmol-1K-1Substance Fe2O3s Fes H2Ol H2gCpo JK-1mol-1 103 25 75 29

Given, nbsp; Fe_2O_3(s)+3H_2(g)→ 2Fe(s)+3H_2O nbsp; We know, Δ C_p^o=(C_p)_P (C_p)_R ⇒ nbsp;Δ C_p = 2×25+3×75 103 3×29=85 J/molK Again, Δ H_T_2 nbsp;Δ ...

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Standard XII

Physics

Calorimetry

Calculate, th...

Question

Calculate, the $Δ G^{o}$ at $85^{o} C$ for the reaction:
$F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 3 H_{2} O$
The data are: $Δ H_{298}^{o} = - 35 k J / m o l$
$Δ S_{358}^{o} = 0.095 k J m o l^{- 1} K^{- 1}$
$S u b s t a n c e F e_{2} O_{3} (s) F e (s) H_{2} O (l) H_{2} (g)$
$C_{p}^{o} (J K^{- 1} m o l - 1) 103257529$

- 63.91 k J m o l^{- 1}

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- 63.91 k J m o l^{- 1}

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- 4.11 k J m o l^{- 1}

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4.11 k J m o l^{- 1}

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Solution

The correct option is A $- 63.91 k J m o l^{- 1}$
Given,
$F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 3 H_{2} O$
We know, $Δ C_{p}^{o} = (C_{p})_{P} - (C_{p})_{R}$
$\Rightarrow Δ C_{p} = 2 \times 25 + 3 \times 75 - 103 - 3 \times 29 = 85 J / m o l K$

Again, $\frac{Δ H_{T_{2}} - Δ H_{T_{1}}}{T_{2} - T_{1}} = Δ C_{p}$
$\Rightarrow \frac{Δ H_{358} - (- 35)}{358 - 298} = 85 \times 10^{- 3}$
$\Rightarrow Δ H_{358}^{o} = - 29.9 k J m o l^{- 1}$

We know, $Δ G^{o} = Δ H^{o} - T Δ S^{o}$
$= - 29.9 - 358 \times 0.095$ $= - 63.91 k J m o l^{- 1}$

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Similar questions

Q. Calculate, the

Δ G^{o}

85^{o} C

for the reaction:

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

The data are:

Δ H_{298}^{o} = - 35 k J / m o l

Δ S_{358}^{o} = 0.095 k J m o l^{- 1} K^{- 1}

S u b s t a n c e F e_{2} O_{3} (s) F e (s) H_{2} O (l) H_{2} (g)

C_{p}^{o} (J K^{- 1} m o l - 1) 103257529

Q. The

Δ H

value at 360 K for the reaction:

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

, given that

Δ H_{300} = - 33.29 k J m o l^{- 1}

and

C_{p}

for

F e_{2} O_{3 (s)}, F e_{(s)}, H_{2} O (l)

and

H_{2 (g)}

are

103.8, 25.1, 75.3

and

28.8

J K^{- 1} m o l^{- 1}

, is:

Q. Calculate

Δ H

358 K

for the reaction,

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

Given that ,

Δ H_{298} = - 33.29 k J m o l^{- 1}

and

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

and

H_{2} (g)

are

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

Q. Calculate

Δ H

358 K

for the reaction,

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

Given that ,

Δ H_{298} = - 33.29 k J m o l^{- 1}

and

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

and

H_{2} (g)

are

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

Q. Calculate standard entropy change in the reaction

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

Given :

S_{m}^{0} (F e_{2} O_{3}, S) = 87.4, S_{m}^{0} (F e, S) = 27.3, S_{m}^{0} (H_{2}, g) = 130.7, S_{m}^{0} (H_{2} O, l) = 69.9 J K^{- 1} m o l^{- 1}

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Calculate, the ΔGo at 85 oC for the reaction: Fe2O3(s)+3H2(g)→2Fe(s)+3H2O The data are: ΔHo298=−35 kJ/mol ΔSo358=0.095 kJmol−1K−1 Substance Fe2O3(s) Fe(s) H2O(l) H2(g) Cop (JK−1mol−1) 103 25 75 29