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Standard XII

Chemistry

Characteristics of Equilibrium Constant

Calculate the...

Question

Calculate the entropy change for the following reaction:
$C a {C O}_{3} (s) ⟶ C a O (s) + C O_{2} (g)$
$S^{o} (J K^{- 1} m o l^{- 1})$ $92.9$ $39.7$ $213.6$

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

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+ 160.4 J K^{- 1} {m o l}^{- 1}

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

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+ 260.4 J K^{- 1} {m o l}^{- 1}

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Solution

The correct option is B $+ 160.4 J K^{- 1} {m o l}^{- 1}$
$Δ S = e n t r o p y o f p r o d u c t - e n t r o p y o f r e a c t a n t$
$Δ S = 39.7 + 213.6 - 92.9 = 160.4 J K^{- 1} m o l^{- 1}$

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Similar questions

Q. Following data is given for the reaction:

C a C O_{3} (s) \to C a O (s) + C O_{2} (g)

Δ_{f} H^{⊝} [C a O (s)] = - 635.1 k J

m o l^{- 1}

Δ_{f} H^{⊝} [C O_{2} (g)] = - 393.5 k J

m o l^{- 1}

Δ_{f} H^{⊝} [C a C O_{3} (s)] = - 1206.9 k J

m o l^{- 1}

Predict the effect of temperature on the equilibrium constant of the above reaction.

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. 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:

Q. Calculate

Δ G^{o}

for the following reaction:

C O (g) + \frac{1}{2} O_{2} (g) ⟶ {C O}_{2} (g); Δ H^{o} = - 282.84 k J

Given,

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

Q. Calculate the entropy change for the following reaction,

H_{2} (g) + C l_{2} (g) ⟶ 2 H C l (g)

298 K

. Given that,

S_{H_{2}}^{⊖} = 131 J K^{- 1} m o l^{- 1}

S_{C l_{2}}^{⊖} = 223 J K^{- 1} m o l^{- 1}

and

S_{H C l}^{⊖} = 187 J K^{- 1} m o l^{- 1}

Calculate the entropy change for the following reaction: CaCO3(s)⟶CaO(s)+CO2(g)So (JK−1mol−1) 92.9 39.7 213.6

The correct option is B +160.4 JK−1mol−1ΔS=entropyofproduct−entropyofreactantΔS=39.7+213.6−92.9=160.4JK−1mol−1

Calculate the entropy change for the following reaction:
$C a {C O}_{3} (s) ⟶ C a O (s) + C O_{2} (g)$
$S^{o} (J K^{- 1} m o l^{- 1})$ $92.9$ $39.7$ $213.6$

The correct option is B $+ 160.4 J K^{- 1} {m o l}^{- 1}$
$Δ S = e n t r o p y o f p r o d u c t - e n t r o p y o f r e a c t a n t$
$Δ S = 39.7 + 213.6 - 92.9 = 160.4 J K^{- 1} m o l^{- 1}$