Calculate the change in entropy for the following reaction nbsp- 2CO-g- - O-2-g- - 2CO-2-g- Given S-0-CO-2-213-6JK-1mol-1-S-0-CO-297-6JK-1mol-1 respectively- nbsp- None of these-185-03 JK-1mol-1-143-03 JK-1mol-1-173-03 JK-1mol-1-

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Standard XII

Chemistry

Entropy

Calculate the...

Question

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.

-185.03 $J K^{- 1} m o l^{- 1}$

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-143.03 $J K^{- 1} m o l^{- 1}$

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-173.03 $J K^{- 1} m o l^{- 1}$

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None of these

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Solution

The correct option is C
-173.03 $J K^{- 1} m o l^{- 1}$

Since, entropy is a state function,
$△$ S = $△ S_{f i n a l} - S_{i n i t i a l}$
For chemical reactions,
$△$ S = $△$ S (products) - $△$ S (reactants)
Therefore,
$= [2 \times S^{⊖} C O_{2} (g) - 2 \times S^{⊖} C O + S^{⊖} O_{2}$
$= (213.6 \times 2) - (2 \times 197.6 + 205.03)$
$= 427.2 - 600.23 = - 173.03 J K^{- 1} m o l^{- 1}$

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Similar questions

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.

Q. Calculate

△ G^{o}

for the following reaction at

298 K

C O (g) + (\frac{1}{2}) O_{2} (g) \to C O_{2} (g); △ H^{o} = - 282.84 kJ/mol

Given,

S_{C O_{2} (g)}^{o} = 213.8 {J K}^{- 1} {mol}^{- 1}, S_{C O (g)}^{o} = 197.9 {J K}^{- 1} {mol}^{- 1}, S_{O_{2}}^{o} = 205.0 {J K}^{- 1} {mol}^{- 1}

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

)

Q. The entropy

(S^{o})

of the following substances are:

C H_{4} (g) = 186.2 J K^{- 1} m o l^{- 1}

O_{2} (g) = 205.0 J K^{- 1} m o l^{- 1}

C O_{2} (g) = 213.6 J K^{- 1} m o l^{- 1}

H_{2} O (l) = 69.9 J K^{- 1} m o l^{- 1}

The entropy change

(Δ S^{o})

for the reaction

C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)

is:

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Calculate the change in entropy for the following reaction 2CO(g)+O2(g)→2CO2(g). Given S0CO2=213.6JK−1mol−1,S0CO=297.6JK−1mol−1 respectively.

The correct option is C -173.03 JK−1mol−1 Since, entropy is a state function, △S = △Sfinal−Sinitial For chemical reactions, △S = △S (products) - △S (reactants) Therefore, =[2×S⊖CO2(g)−2×S⊖CO+S⊖O2 =(213.6×2)−(2×197.6+205.03) =427.2−600.23=−173.03 JK−1mol−1

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.