Calculate $△ G^{⊖}$ for the following reaction:
$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)$ ; $△ H^{⊖}$ = -282.84 kJ
Given,
$S^{⊖} C o_{2} = 213.8 J$ $K^{- 1} m o l^{- 1}$ , $S^{⊖} C O (g)$ = 197.9 J $K^{- 1} m o l^{- 1}$ ,
$S^{⊖} C o_{2} = 205.0 J$ $K^{- 1} m o l^{- 1}$

Q. For the reaction:

A (s) ⟶ B (s) + C (s) J K^{- 1} {m o l}^{- 1}

Calculate the entropy change at

298 K

and

1

atm if absolute entropies (in

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

) are:

A = 130, B = 203, C = 152

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For a reaction: X→Y+Z Absolute entropies of X,Y,Z are 120 J K−1 mol−1 ,213.8 JK−1 mol−1 and 197.9 J K−1 mol−1 respectively.What will be the entropy change of the reaction at 298 K and 1 atm ?

The correct option is A 291.7 J K−1ΔS0=∑S0P−∑S0R =(213.8+197.9)−120 =291.7 JK−1Hence, the correct answer is option A.

For a reaction: $X \to Y + Z$
Absolute entropies of X,Y,Z are 120 J $K^{- 1} m o l^{- 1}$ ,213.8 J $K^{- 1} m o l^{- 1}$ and 197.9 J $K^{- 1} m o l^{- 1}$ respectively.
What will be the entropy change of the reaction at 298 K and 1 atm ?

The correct option is A 291.7 J $K^{- 1}$
$Δ S^{0} = \sum S_{P}^{0} - \sum S_{R}^{0}$

$= (213.8 + 197.9) - 120$

$= 291.7 J K^{- 1}$

Hence, the correct answer is option $A$ .