QBManager

Thermodynamics

272745 For which of the process, $Δ S$ is negative?

1

H_{2}

(g)

\to 2 H

(g)

2

N_{2}

(g)

(1 atm) \to N_{2}

(g) (8 atm)

3

2 {SO}_{3}

(g)

\to 2 {SO}_{2}

(g)

+ O_{2}

(g)

4

C_{(diamond)} \to C_{(graplite)}

Explanation:

The higher is the pressure lower become entropy. When the pressure increase on the gases molecule kinetic energy of the particle decrease with respect to the pressure. This is due to the particle comes closer and intermolecular distance decreases the entropy.
Entropy is a state function that is often referred to as the state of disorder of a system.
- $H_{2} (g) \to 2 H (g)$ entropy increases because no. of atoms increases.
- $N_{2} (g, 1 atm) \to N_{2} (g, 8 atm)$ entropy decreases as pressure increase and randomness decreases.
- $2 {SO}_{3} (g) \to 2 {SO}_{2} (g) + O_{2} (g)$ entropy increases because no. of atoms increases.
- $C_{(dimond)} \to C_{(graphite)}$ entropy increases as the compactness of system decreases and randomness increases.

BITSAT 2008

Thermodynamics

272746 For a given reaction, $Δ H = 35.5 kJ {mol}^{- 1}$ and $Δ S = 83.6 {JK}^{- 1} {mol}^{- 1}$ . The reaction is spontaneous at : (Assume that $Δ H$ and $Δ S$ do not vary with temperature)

1

T > 425 K

2 All temperature

3

T > 298 K

4

T < 425 K

Explanation:

Given that
$Δ H = 35.5 kJ / mol = 35500 J / mol$
$Δ S = 83.6 J / K / mol$
$We know$ that,
$Δ G = Δ H - T Δ S$
For spontaneous $Δ G < 0$ $Δ H - T Δ S < 0$
$T > \frac{Δ H}{Δ S}$
$T > \frac{35500}{83.6}$
$T > 425 K$
So, the given reaction will be spontaneous at $T > 425 K$ .

BITSAT 2017

Thermodynamics

272754 The values of $Δ H$ and $Δ S$ for a reaction are respectively $30 kJ {mol}^{- 1}$ and $100 {JK}^{- 1} {mol}^{- 1}$ . Then the temperature above which the reaction will become spontaneous is

1

300 K

2

30 K

3

100 K

4

300^{\circ} C

Explanation:

Given that $Δ H = 30 kJ {mol}^{- 1} = 30 \times 10^{3} {Jmol}^{- 1}$
$Δ S = 100 {JK}^{- 1}$
$Δ G = Δ H - T Δ S$
For spontaneous reaction, $Δ G < 0$
$Δ H - T Δ S < 0$
$(30 \times 10^{3}) - T \times (100) < 0$
$30000 < 100 T$
$300 < T$
$T > 300 K$

J and K CET-(2011)

Thermodynamics

272760 When the same quantity of heat is absorbed by a system at two different temperatures $T_{1}$ and $T_{2}$ , such that $T_{1} > T_{2}$ , change in entropies are $Δ S_{1}$ and $Δ S_{2}$ respectively. Then

1

Δ S_{1} < Δ S_{2}

2

Δ S_{1} = Δ S_{2}

3

S_{1} > S_{2}

4

Δ S_{2} < Δ S_{1}

Explanation:

Entropy change occur due to transfer of heat. $Δ S = q / T [q =$ constant $]$
So, $Δ S \propto \frac{1}{T}$
Given, $T_{1} > T_{2}$
So, actual relation between the entropies will be
$(Δ S_{2} > Δ S_{1})$

Karnataka-CET-2020

Thermodynamics

272745 For which of the process, $Δ S$ is negative?

1

H_{2}

(g)

\to 2 H

(g)

2

N_{2}

(g)

(1 atm) \to N_{2}

(g) (8 atm)

3

2 {SO}_{3}

(g)

\to 2 {SO}_{2}

(g)

+ O_{2}

(g)

4

C_{(diamond)} \to C_{(graplite)}

Explanation:

The higher is the pressure lower become entropy. When the pressure increase on the gases molecule kinetic energy of the particle decrease with respect to the pressure. This is due to the particle comes closer and intermolecular distance decreases the entropy.
Entropy is a state function that is often referred to as the state of disorder of a system.
- $H_{2} (g) \to 2 H (g)$ entropy increases because no. of atoms increases.
- $N_{2} (g, 1 atm) \to N_{2} (g, 8 atm)$ entropy decreases as pressure increase and randomness decreases.
- $2 {SO}_{3} (g) \to 2 {SO}_{2} (g) + O_{2} (g)$ entropy increases because no. of atoms increases.
- $C_{(dimond)} \to C_{(graphite)}$ entropy increases as the compactness of system decreases and randomness increases.

BITSAT 2008

Thermodynamics

272746 For a given reaction, $Δ H = 35.5 kJ {mol}^{- 1}$ and $Δ S = 83.6 {JK}^{- 1} {mol}^{- 1}$ . The reaction is spontaneous at : (Assume that $Δ H$ and $Δ S$ do not vary with temperature)

1

T > 425 K

2 All temperature

3

T > 298 K

4

T < 425 K

Explanation:

Given that
$Δ H = 35.5 kJ / mol = 35500 J / mol$
$Δ S = 83.6 J / K / mol$
$We know$ that,
$Δ G = Δ H - T Δ S$
For spontaneous $Δ G < 0$ $Δ H - T Δ S < 0$
$T > \frac{Δ H}{Δ S}$
$T > \frac{35500}{83.6}$
$T > 425 K$
So, the given reaction will be spontaneous at $T > 425 K$ .

BITSAT 2017

Thermodynamics

272754 The values of $Δ H$ and $Δ S$ for a reaction are respectively $30 kJ {mol}^{- 1}$ and $100 {JK}^{- 1} {mol}^{- 1}$ . Then the temperature above which the reaction will become spontaneous is

1

300 K

2

30 K

3

100 K

4

300^{\circ} C

Explanation:

Given that $Δ H = 30 kJ {mol}^{- 1} = 30 \times 10^{3} {Jmol}^{- 1}$
$Δ S = 100 {JK}^{- 1}$
$Δ G = Δ H - T Δ S$
For spontaneous reaction, $Δ G < 0$
$Δ H - T Δ S < 0$
$(30 \times 10^{3}) - T \times (100) < 0$
$30000 < 100 T$
$300 < T$
$T > 300 K$

J and K CET-(2011)

Thermodynamics

272760 When the same quantity of heat is absorbed by a system at two different temperatures $T_{1}$ and $T_{2}$ , such that $T_{1} > T_{2}$ , change in entropies are $Δ S_{1}$ and $Δ S_{2}$ respectively. Then

1

Δ S_{1} < Δ S_{2}

2

Δ S_{1} = Δ S_{2}

3

S_{1} > S_{2}

4

Δ S_{2} < Δ S_{1}

Explanation:

Entropy change occur due to transfer of heat. $Δ S = q / T [q =$ constant $]$
So, $Δ S \propto \frac{1}{T}$
Given, $T_{1} > T_{2}$
So, actual relation between the entropies will be
$(Δ S_{2} > Δ S_{1})$

Karnataka-CET-2020

Thermodynamics

272745 For which of the process, $Δ S$ is negative?

1

H_{2}

(g)

\to 2 H

(g)

2

N_{2}

(g)

(1 atm) \to N_{2}

(g) (8 atm)

3

2 {SO}_{3}

(g)

\to 2 {SO}_{2}

(g)

+ O_{2}

(g)

4

C_{(diamond)} \to C_{(graplite)}

Explanation:

The higher is the pressure lower become entropy. When the pressure increase on the gases molecule kinetic energy of the particle decrease with respect to the pressure. This is due to the particle comes closer and intermolecular distance decreases the entropy.
Entropy is a state function that is often referred to as the state of disorder of a system.
- $H_{2} (g) \to 2 H (g)$ entropy increases because no. of atoms increases.
- $N_{2} (g, 1 atm) \to N_{2} (g, 8 atm)$ entropy decreases as pressure increase and randomness decreases.
- $2 {SO}_{3} (g) \to 2 {SO}_{2} (g) + O_{2} (g)$ entropy increases because no. of atoms increases.
- $C_{(dimond)} \to C_{(graphite)}$ entropy increases as the compactness of system decreases and randomness increases.

BITSAT 2008

Thermodynamics

272746 For a given reaction, $Δ H = 35.5 kJ {mol}^{- 1}$ and $Δ S = 83.6 {JK}^{- 1} {mol}^{- 1}$ . The reaction is spontaneous at : (Assume that $Δ H$ and $Δ S$ do not vary with temperature)

1

T > 425 K

2 All temperature

3

T > 298 K

4

T < 425 K

Explanation:

Given that
$Δ H = 35.5 kJ / mol = 35500 J / mol$
$Δ S = 83.6 J / K / mol$
$We know$ that,
$Δ G = Δ H - T Δ S$
For spontaneous $Δ G < 0$ $Δ H - T Δ S < 0$
$T > \frac{Δ H}{Δ S}$
$T > \frac{35500}{83.6}$
$T > 425 K$
So, the given reaction will be spontaneous at $T > 425 K$ .

BITSAT 2017

Thermodynamics

272754 The values of $Δ H$ and $Δ S$ for a reaction are respectively $30 kJ {mol}^{- 1}$ and $100 {JK}^{- 1} {mol}^{- 1}$ . Then the temperature above which the reaction will become spontaneous is

1

300 K

2

30 K

3

100 K

4

300^{\circ} C

Explanation:

Given that $Δ H = 30 kJ {mol}^{- 1} = 30 \times 10^{3} {Jmol}^{- 1}$
$Δ S = 100 {JK}^{- 1}$
$Δ G = Δ H - T Δ S$
For spontaneous reaction, $Δ G < 0$
$Δ H - T Δ S < 0$
$(30 \times 10^{3}) - T \times (100) < 0$
$30000 < 100 T$
$300 < T$
$T > 300 K$

J and K CET-(2011)

Thermodynamics

272760 When the same quantity of heat is absorbed by a system at two different temperatures $T_{1}$ and $T_{2}$ , such that $T_{1} > T_{2}$ , change in entropies are $Δ S_{1}$ and $Δ S_{2}$ respectively. Then

1

Δ S_{1} < Δ S_{2}

2

Δ S_{1} = Δ S_{2}

3

S_{1} > S_{2}

4

Δ S_{2} < Δ S_{1}

Explanation:

Entropy change occur due to transfer of heat. $Δ S = q / T [q =$ constant $]$
So, $Δ S \propto \frac{1}{T}$
Given, $T_{1} > T_{2}$
So, actual relation between the entropies will be
$(Δ S_{2} > Δ S_{1})$

Karnataka-CET-2020

Thermodynamics

272745 For which of the process, $Δ S$ is negative?

1

H_{2}

(g)

\to 2 H

(g)

2

N_{2}

(g)

(1 atm) \to N_{2}

(g) (8 atm)

3

2 {SO}_{3}

(g)

\to 2 {SO}_{2}

(g)

+ O_{2}

(g)

4

C_{(diamond)} \to C_{(graplite)}

Explanation:

The higher is the pressure lower become entropy. When the pressure increase on the gases molecule kinetic energy of the particle decrease with respect to the pressure. This is due to the particle comes closer and intermolecular distance decreases the entropy.
Entropy is a state function that is often referred to as the state of disorder of a system.
- $H_{2} (g) \to 2 H (g)$ entropy increases because no. of atoms increases.
- $N_{2} (g, 1 atm) \to N_{2} (g, 8 atm)$ entropy decreases as pressure increase and randomness decreases.
- $2 {SO}_{3} (g) \to 2 {SO}_{2} (g) + O_{2} (g)$ entropy increases because no. of atoms increases.
- $C_{(dimond)} \to C_{(graphite)}$ entropy increases as the compactness of system decreases and randomness increases.

BITSAT 2008

Thermodynamics

272746 For a given reaction, $Δ H = 35.5 kJ {mol}^{- 1}$ and $Δ S = 83.6 {JK}^{- 1} {mol}^{- 1}$ . The reaction is spontaneous at : (Assume that $Δ H$ and $Δ S$ do not vary with temperature)

1

T > 425 K

2 All temperature

3

T > 298 K

4

T < 425 K

Explanation:

Given that
$Δ H = 35.5 kJ / mol = 35500 J / mol$
$Δ S = 83.6 J / K / mol$
$We know$ that,
$Δ G = Δ H - T Δ S$
For spontaneous $Δ G < 0$ $Δ H - T Δ S < 0$
$T > \frac{Δ H}{Δ S}$
$T > \frac{35500}{83.6}$
$T > 425 K$
So, the given reaction will be spontaneous at $T > 425 K$ .

BITSAT 2017

Thermodynamics

272754 The values of $Δ H$ and $Δ S$ for a reaction are respectively $30 kJ {mol}^{- 1}$ and $100 {JK}^{- 1} {mol}^{- 1}$ . Then the temperature above which the reaction will become spontaneous is

1

300 K

2

30 K

3

100 K

4

300^{\circ} C

Explanation:

Given that $Δ H = 30 kJ {mol}^{- 1} = 30 \times 10^{3} {Jmol}^{- 1}$
$Δ S = 100 {JK}^{- 1}$
$Δ G = Δ H - T Δ S$
For spontaneous reaction, $Δ G < 0$
$Δ H - T Δ S < 0$
$(30 \times 10^{3}) - T \times (100) < 0$
$30000 < 100 T$
$300 < T$
$T > 300 K$

J and K CET-(2011)

Thermodynamics

272760 When the same quantity of heat is absorbed by a system at two different temperatures $T_{1}$ and $T_{2}$ , such that $T_{1} > T_{2}$ , change in entropies are $Δ S_{1}$ and $Δ S_{2}$ respectively. Then

1

Δ S_{1} < Δ S_{2}

2

Δ S_{1} = Δ S_{2}

3

S_{1} > S_{2}

4

Δ S_{2} < Δ S_{1}

Explanation:

Entropy change occur due to transfer of heat. $Δ S = q / T [q =$ constant $]$
So, $Δ S \propto \frac{1}{T}$
Given, $T_{1} > T_{2}$
So, actual relation between the entropies will be
$(Δ S_{2} > Δ S_{1})$

Karnataka-CET-2020