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CHXI06:THERMODYNAMICS

369274 Consider the reactions at $300 K$
(a) $4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g); Δ H = - 111 kJ$
(b) $4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (s); Δ H = - 219 kJ$
Thus entropy change for the sublimation of one mole $N_{2} O_{5} (s)$ is :

1

180 J {mol}^{- 1} K^{- 1}

2

360 J {mol}^{- 1} K^{- 1}

3

108 J {mol}^{- 1} K^{- 1}

4

54 J {mol}^{- 1} K^{- 1}

Explanation:

Reverse the reaction
(b) and add reaction
(a) to get $2 \underset{(s)}{N_{2} O_{5}} \to 2 \underset{(g)}{N_{2} O_{5}};$
$Δ H = + 219 - 111 = + 108 kJ$ for 2 moles
$∴$ for $1 mole = + 54 kJ = 54000 J$
$Δ S_{sub} = \frac{Δ_{sub} H}{T} = \frac{54000}{300} = 180 J {mol}^{- 1} K^{- 1}$

CHXI06:THERMODYNAMICS

369275 In which of the following process entropy increases?
(I) Rusting of iron
(II) Vapourisation of Camphor
(III) Crystallisation of sugar from syrup
(IV) Atomisation of dihydrogen

1 I and II

2 II and III

3 II and IV

4 Only IV

Explanation:

(I) Rusting of iron
$Fe (solid) + O_{2} (gas) ⟶ {Fe}_{2} O_{3} (solid)$
Entropy decreases during above process as gas convert to solid.
(II) During vapourisation of camphor entropy increases as solid converts to gas.
(III) During crystallisation of sugar from syrup cause entropy decreases as solid substance formed.
(IV) During atomisation of dihydrogen entropy decreases as number of particles decreases.

CHXI06:THERMODYNAMICS

369276 For the reaction, $2 {SO}_{2 (g)} + O_{2 (g)} \to 2 {SO}_{3 (g)}$ , the entropy

1 Increases

2 Decreases

3 Remains unchanged

4 Change cannot be predicted

CHXI06:THERMODYNAMICS

369277 Assertion :
Chlorine when solidified does not have zero entropy even at absolute zero.
Reason :
Chlorine is a pungent smelling gas and it is difficult to solidify it.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

For substance to have zero entropy zero, it must be a perfectly substance. Chlorine is a diatomic molecule and does not form a perfect crystal structure when solidified. This means that even when chlorine is solidified, it does not achieve the perfect order required for its entropy to be zero. Therefore, the assertion is correct. ${Cl}_{2}$ has a pungent smell and is difficult to be solidified. But, it does not explain why its entropy does not reach zero when solidified. So, option (2) is correct.

CHXI06:THERMODYNAMICS

369274 Consider the reactions at $300 K$
(a) $4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g); Δ H = - 111 kJ$
(b) $4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (s); Δ H = - 219 kJ$
Thus entropy change for the sublimation of one mole $N_{2} O_{5} (s)$ is :

1

180 J {mol}^{- 1} K^{- 1}

2

360 J {mol}^{- 1} K^{- 1}

3

108 J {mol}^{- 1} K^{- 1}

4

54 J {mol}^{- 1} K^{- 1}

Explanation:

Reverse the reaction
(b) and add reaction
(a) to get $2 \underset{(s)}{N_{2} O_{5}} \to 2 \underset{(g)}{N_{2} O_{5}};$
$Δ H = + 219 - 111 = + 108 kJ$ for 2 moles
$∴$ for $1 mole = + 54 kJ = 54000 J$
$Δ S_{sub} = \frac{Δ_{sub} H}{T} = \frac{54000}{300} = 180 J {mol}^{- 1} K^{- 1}$

CHXI06:THERMODYNAMICS

369275 In which of the following process entropy increases?
(I) Rusting of iron
(II) Vapourisation of Camphor
(III) Crystallisation of sugar from syrup
(IV) Atomisation of dihydrogen

1 I and II

2 II and III

3 II and IV

4 Only IV

Explanation:

(I) Rusting of iron
$Fe (solid) + O_{2} (gas) ⟶ {Fe}_{2} O_{3} (solid)$
Entropy decreases during above process as gas convert to solid.
(II) During vapourisation of camphor entropy increases as solid converts to gas.
(III) During crystallisation of sugar from syrup cause entropy decreases as solid substance formed.
(IV) During atomisation of dihydrogen entropy decreases as number of particles decreases.

CHXI06:THERMODYNAMICS

369276 For the reaction, $2 {SO}_{2 (g)} + O_{2 (g)} \to 2 {SO}_{3 (g)}$ , the entropy

1 Increases

2 Decreases

3 Remains unchanged

4 Change cannot be predicted

CHXI06:THERMODYNAMICS

369277 Assertion :
Chlorine when solidified does not have zero entropy even at absolute zero.
Reason :
Chlorine is a pungent smelling gas and it is difficult to solidify it.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

For substance to have zero entropy zero, it must be a perfectly substance. Chlorine is a diatomic molecule and does not form a perfect crystal structure when solidified. This means that even when chlorine is solidified, it does not achieve the perfect order required for its entropy to be zero. Therefore, the assertion is correct. ${Cl}_{2}$ has a pungent smell and is difficult to be solidified. But, it does not explain why its entropy does not reach zero when solidified. So, option (2) is correct.

CHXI06:THERMODYNAMICS

369274 Consider the reactions at $300 K$
(a) $4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g); Δ H = - 111 kJ$
(b) $4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (s); Δ H = - 219 kJ$
Thus entropy change for the sublimation of one mole $N_{2} O_{5} (s)$ is :

1

180 J {mol}^{- 1} K^{- 1}

2

360 J {mol}^{- 1} K^{- 1}

3

108 J {mol}^{- 1} K^{- 1}

4

54 J {mol}^{- 1} K^{- 1}

Explanation:

Reverse the reaction
(b) and add reaction
(a) to get $2 \underset{(s)}{N_{2} O_{5}} \to 2 \underset{(g)}{N_{2} O_{5}};$
$Δ H = + 219 - 111 = + 108 kJ$ for 2 moles
$∴$ for $1 mole = + 54 kJ = 54000 J$
$Δ S_{sub} = \frac{Δ_{sub} H}{T} = \frac{54000}{300} = 180 J {mol}^{- 1} K^{- 1}$

CHXI06:THERMODYNAMICS

369275 In which of the following process entropy increases?
(I) Rusting of iron
(II) Vapourisation of Camphor
(III) Crystallisation of sugar from syrup
(IV) Atomisation of dihydrogen

1 I and II

2 II and III

3 II and IV

4 Only IV

Explanation:

(I) Rusting of iron
$Fe (solid) + O_{2} (gas) ⟶ {Fe}_{2} O_{3} (solid)$
Entropy decreases during above process as gas convert to solid.
(II) During vapourisation of camphor entropy increases as solid converts to gas.
(III) During crystallisation of sugar from syrup cause entropy decreases as solid substance formed.
(IV) During atomisation of dihydrogen entropy decreases as number of particles decreases.

CHXI06:THERMODYNAMICS

369276 For the reaction, $2 {SO}_{2 (g)} + O_{2 (g)} \to 2 {SO}_{3 (g)}$ , the entropy

1 Increases

2 Decreases

3 Remains unchanged

4 Change cannot be predicted

CHXI06:THERMODYNAMICS

369277 Assertion :
Chlorine when solidified does not have zero entropy even at absolute zero.
Reason :
Chlorine is a pungent smelling gas and it is difficult to solidify it.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

For substance to have zero entropy zero, it must be a perfectly substance. Chlorine is a diatomic molecule and does not form a perfect crystal structure when solidified. This means that even when chlorine is solidified, it does not achieve the perfect order required for its entropy to be zero. Therefore, the assertion is correct. ${Cl}_{2}$ has a pungent smell and is difficult to be solidified. But, it does not explain why its entropy does not reach zero when solidified. So, option (2) is correct.

CHXI06:THERMODYNAMICS

369274 Consider the reactions at $300 K$
(a) $4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (g); Δ H = - 111 kJ$
(b) $4 N O_{2} (g) + O_{2} (g) \to 2 N_{2} O_{5} (s); Δ H = - 219 kJ$
Thus entropy change for the sublimation of one mole $N_{2} O_{5} (s)$ is :

1

180 J {mol}^{- 1} K^{- 1}

2

360 J {mol}^{- 1} K^{- 1}

3

108 J {mol}^{- 1} K^{- 1}

4

54 J {mol}^{- 1} K^{- 1}

Explanation:

Reverse the reaction
(b) and add reaction
(a) to get $2 \underset{(s)}{N_{2} O_{5}} \to 2 \underset{(g)}{N_{2} O_{5}};$
$Δ H = + 219 - 111 = + 108 kJ$ for 2 moles
$∴$ for $1 mole = + 54 kJ = 54000 J$
$Δ S_{sub} = \frac{Δ_{sub} H}{T} = \frac{54000}{300} = 180 J {mol}^{- 1} K^{- 1}$

CHXI06:THERMODYNAMICS

369275 In which of the following process entropy increases?
(I) Rusting of iron
(II) Vapourisation of Camphor
(III) Crystallisation of sugar from syrup
(IV) Atomisation of dihydrogen

1 I and II

2 II and III

3 II and IV

4 Only IV

Explanation:

(I) Rusting of iron
$Fe (solid) + O_{2} (gas) ⟶ {Fe}_{2} O_{3} (solid)$
Entropy decreases during above process as gas convert to solid.
(II) During vapourisation of camphor entropy increases as solid converts to gas.
(III) During crystallisation of sugar from syrup cause entropy decreases as solid substance formed.
(IV) During atomisation of dihydrogen entropy decreases as number of particles decreases.

CHXI06:THERMODYNAMICS

369276 For the reaction, $2 {SO}_{2 (g)} + O_{2 (g)} \to 2 {SO}_{3 (g)}$ , the entropy

1 Increases

2 Decreases

3 Remains unchanged

4 Change cannot be predicted

CHXI06:THERMODYNAMICS

369277 Assertion :
Chlorine when solidified does not have zero entropy even at absolute zero.
Reason :
Chlorine is a pungent smelling gas and it is difficult to solidify it.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

For substance to have zero entropy zero, it must be a perfectly substance. Chlorine is a diatomic molecule and does not form a perfect crystal structure when solidified. This means that even when chlorine is solidified, it does not achieve the perfect order required for its entropy to be zero. Therefore, the assertion is correct. ${Cl}_{2}$ has a pungent smell and is difficult to be solidified. But, it does not explain why its entropy does not reach zero when solidified. So, option (2) is correct.