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Molar and Specific Heat Capacity

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Question

Calculate the entropy change for the following reaction, $H_{2} (g) + C l_{2} (g) ⟶ 2 H C l (g)$ at $298 K$ . 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 $S_{H C l}^{⊖} = 187 J K^{- 1} m o l^{- 1}$ .

Δ S = 20 J K^{- 1} m o l^{- 1}

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Δ S = - 20 J K^{- 1} m o l^{- 1}

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Δ S = 40 J K^{- 1} m o l^{- 1}

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None of these

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Solution

The correct option is D $Δ S = 20 J K^{- 1} m o l^{- 1}$
$Δ S^{⊖}$ $= \sum S^{⊖} (P r o d u c t) - \sum S^{⊖} (R e a c t a n t) Δ S^{⊖}$

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

$Δ S^{⊖}$ $= 2 \times 187 - (131 + 223)$

$Δ S^{⊖}$ $= 2 \times 187 - (131 + 223)$

$Δ S^{⊖}$ $= 374 - 131 - 223$

$Δ S^{⊖}$ $= 374 - 131 - 223$

$Δ S^{⊖}$ $= 20 J K^{- 1} m o l^{- 1}$

$Δ S^{⊖}$ $= 20 J K^{- 1} m o l^{- 1}$

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Similar questions

If S⁰ for H₂, Cl₂ and HCL are 0.13, 0.22 and 0.19 kJK^-1mol^-1 respectively. The total change in standard entropy for th2e reaction, H₂+Cl₂→2HCl

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.

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

)

The entropy values in

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

H_{2} (g) = 130.6, {C l}_{2} (g) = 233

and

H C l (g) = 186.7

298 K

and

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atm pressure. Then entropy change for the reaction

H_{2} (g) + {C l}_{2} (g) \to 2 H C l (g)

is:

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Calculate the entropy change for the following reaction, H2(g)+Cl2(g)⟶2HCl(g) at 298K. Given that, S⊖H2=131JK−1mol−1, S⊖Cl2=223JK−1mol−1 and S⊖HCl=187JK−1mol−1.

The correct option is D ΔS=20 JK−1mol−1ΔS⊖ =∑S⊖(Product)−∑S⊖(Reactant)ΔS⊖ ΔS⊖ =∑S⊖(Product)−∑S⊖(Reactant) ΔS⊖ =2×187−(131+223)ΔS⊖ =2×187−(131+223) ΔS⊖ =374−131−223ΔS⊖ =374−131−223 ΔS⊖ =20JK−1mol−1ΔS⊖ =20JK−1mol−1

Calculate the entropy change for the following reaction, $H_{2} (g) + C l_{2} (g) ⟶ 2 H C l (g)$ at $298 K$ . 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 $S_{H C l}^{⊖} = 187 J K^{- 1} m o l^{- 1}$ .