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Standard XII

Chemistry

Haber Bosch Process

If S0 for H2,...

Question

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

30 J K^-1mol^-1

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40 J K^-1mol^-1

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60 J K^-1mol^-1

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20 J K^-1mol^-1

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Solution

The correct option is A 30 J K^-1mol^-1
∆S = S_p - S_R
(2×0.19) - 0.13 - 0.22
= 0.03 KJK^-1mol^-1
=30 K^-1mol^-1

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Similar questions

Q. If

S^{0}

for

H_{2}

{C l}_{2}

and HCl are 0.13, 0.22 and 0.19 kJ

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

respectively. The total change in standard entropy for the reaction

H_{2} + {C l}_{2} ⟶ 2 H C l

is:

Q. If

S^{o}

for

H_{2}

C l_{2}

and HCl are 0.13, 0.22 and 0.19 KJ

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

respectively. The total change is standard entropy for the reaction,

H_{2} + C l_{2} \to 2 H C l

is:

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. If

S^{o}

for

H_{2}, {C l}_{2}

and

H C l

are

0.13, 0.22

and

0.19 {k J K}^{- 1}

{m o l}^{- 1}

respectively. The total change in standard entropy for the reaction

H_{2} + {C l}_{2} \to 2 H C l

is:

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

)

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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

The correct option is A 30 J K-1mol-1∆S = Sp - SR(2×0.19) - 0.13 - 0.22 = 0.03 KJK-1mol-1 =30 K-1mol-1

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

The correct option is A 30 J K^-1mol^-1
∆S = S_p - S_R
(2×0.19) - 0.13 - 0.22
= 0.03 KJK^-1mol^-1
=30 K^-1mol^-1