Calucalculate standard entropy change in the reaction F e 2 O 3 s -3 H 2 g - 2 Fe s -2 H 2 O 1Given- Sm0 Fe 2 O 3- S -87-4- S m 0 Fe - S -27-3S m 0 H 2- g -130-7- S m 0 H 2 O - 1-69-9 J K -1 mol 2A- -212-5 JK -1B- -215-2 JK -1C- -120-9 JK -1D- None of these

The correct option is B 215.2 JK 1 Fe_2O_3(s) + 3H_2(g) → 2Fe(s) + 3H_2O(s) Changed balanced equation Δ S = ∑ S^0(Products) ∑ S^0(Reactants) [2S^0_Fe + 2S^0 ...

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Byju's Answer

Standard XII

Chemistry

Electrolysis and Electrolytes

Calucalculate...

Question

Calucalculate standard entropy change in the reaction

$F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 2 H_{2} O (1)$

Given: $S_{m}^{o} (F e_{2} O_{3}, S) = 87.4, S_{m}^{o} (F e, S) = 27.3$

$S_{m}^{o} (H_{2}, g) = 130.7, S_{m}^{o} (H_{2} O, 1) = 69.9 J K^{- 1} m o l^{2}$

-212.5 JK^-1

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-215.2 JK^-1

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-120.9 JK^-1

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None of these

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Solution

The correct option is B
-215.2 JK^-1

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

Changed balanced equation

$Δ S = \sum S^{0} (P r o d u c t s) - \sum S^{0} (R e a c t a n t s)$

$[2 S_{F e}^{0} + 2 S_{H_{2} O}^{0}] - [S_{F e_{2} O_{3}}^{0} + 3 S_{H_{2}}^{0}]$

= $(2 \times 27.3 + 3 \times 69.9) - [87.4 + 3 \times 130.7] = (54.6 + 209.7) - [87.4 + 92.1]$

= $264.3 - 479.5 = - 215.2 J / K / m o l$

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Similar questions

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}

Q. In the reaction

A^{+} + B \to A + B^{+},

there is no entropy change. If enthalpy change is

22 k J / m o l

, calculate

△ G

for the reaction.

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

The reducing agent in the above reaction is:

Q. Given that:

S_{H_{2}}^{\circ} = 131 J K^{- 1} m o l^{- 1} S_{C l_{2}}^{\circ} = 223 J K^{- 1} m o l^{- 1} a n d S_{H C l}^{\circ} = 187 J K^{- 1} m o l^{- 1}

The standard entropy change in formation of

1

mole of

H C l (g)

from

H_{2} (g)

and

C l_{2} (g)

will be:
(Given, reaction for formation of

H C l

H_{2} + C l_{2} \to 2 H C l

)

Calculate the change in entropy for the following reaction
$2 C O (g) + O_{2} (g) \to 2 C O_{2} (g)$ .
Given $S_{C O_{2}}^{0} = 213.6 J K^{- 1} m o l^{- 1}$ , $S_{C O}^{0} = 297.6 J K^{- 1} m o l^{- 1}$ respectively.

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Calucalculate standard entropy change in the reaction Fe2O3(s) + 3H2(g) → 2Fe(s) + 2H2O(1) Given: Som(Fe2O3,S) = 87.4, Som(Fe,S) = 27.3 Som(H2,g) = 130.7, Som(H2O,1)= 69.9 JK−1mol2

Calucalculate standard entropy change in the reaction

$F e_{2} O_{3} (s) + 3 H_{2} (g) \to 2 F e (s) + 2 H_{2} O (1)$

Given: $S_{m}^{o} (F e_{2} O_{3}, S) = 87.4, S_{m}^{o} (F e, S) = 27.3$

$S_{m}^{o} (H_{2}, g) = 130.7, S_{m}^{o} (H_{2} O, 1) = 69.9 J K^{- 1} m o l^{2}$