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Entropy

For a reactio...

Question

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

291.7 J K^{- 1}

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255 J K^{- 1}

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213.8 J K^{- 1}

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257.3 J K^{- 1}

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Solution

The correct option is A $291.7 J K^{- 1}$
Change in entropy for a reaction is given as:
$Δ S =$ Entropy of the product - Entropy of the reactant
So, $Δ S$ = $S_{p r o d u c t} - S_{r e a c t a n t}$
$Δ S$ = $(S_{Y} + S_{Z}) - S_{X}$
= $(213.8 + 197.9) - 120$
$Δ S$ = $291.7 J K^{- 1} m o l^{- 1}$

Theory:
$Δ S = S_{f i n a l} - S_{i n i t i a l}$

$Δ S = \sum S_{p r o d u c t s} - \sum S_{r e a c t a n t s}$

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Similar questions

Q. 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 ?

Q. For a reaction:

X \to Y + Z

The absolute entropies for

X, Y

and

Z

are

120 J K^{- 1} m o l^{- 1}, 213.8 J K^{- 1} m o l^{- 1} a n d 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?

S_{H_{2 (g)}}^{O} = 130.6

K^{- 1}

m o l 6 - 1

;

S_{H_{2} O_{(l)}}^{O} = 69.9

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

S_{O_{2 (g)}}^{O} = 205

K^{- 1}

m o l^{- 1}

, then the absolute entropy change of

H_{2_{(g)}} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)}

is?

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

For a reaction at $25^{\circ} C$ enthalpy change ( $△ H$ ) and entropy change ( $△ S$ ) are -11.7 K J $m o l^{- 1}$ and -105 J $m o l^{- 1}$ $K^{- 1}$ , respectively. It is a spontaneous reaction

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Entropy

Standard XI Chemistry

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For a reaction: X→Y+Z The absolute entropies for X,Y and Z are 120 J K−1mol−1, 213.8 J K−1 mol−1 and 197.9 J K−1mol−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−1Change in entropy for a reaction is given as: ΔS= Entropy of the product - Entropy of the reactant So, ΔS = Sproduct−Sreactant ΔS = (SY+SZ)−SX = (213.8+197.9)−120 ΔS = 291.7 JK−1mol−1 Theory: ΔS=Sfinal−Sinitial ΔS=∑Sproducts−∑Sreactants

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