QBManager

Thermodynamics

272398 For the reaction
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), Δ H$
$= - 285.8 kJ {mol}^{- 1}$
$Δ S = - 0.163 kJ {mol}^{- 1} K^{- 1}$
What is the value of free energy change at $27^{\circ} C$ for the reaction?

1

- 236.9 kJ {mol}^{- 1}

2

- 281.4 kJ {mol}^{- 1}

3

- 334.7 kJ {mol}^{- 1}

4

+ 334.7 kJ {mol}^{- 1}

Explanation:

Given,
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$
$Δ H = - 285.8 kJ / mole$
$Δ S = - 0.163 kJ / mole$
$Δ G = Δ H - T Δ S$
$Δ G = - 285.8 + 300 \times 0.163$
$Δ G = - 236.97 kJ / mole$

Thermodynamics

272399 In a closed insulated container, a liquid is stirred with a paddle to increase its temperature. In this process, which of the following is true?

1

Δ E = W = Q = 0

2

Δ E \neq 0, Q = W = 0

3

Δ E = W \neq 0, Q = 0

4

Δ E = Q \neq 0, W = 0

Explanation:

According to $I I^{st}$ Law of Thermodynamic
$q = Δ E + W$
$E =$ change in internal energy
$W =$ work done on system
Since system is closed and insulated, $q = 0$
Paddle work is done on system.
Hence internal energy of the system is increases.
$∴ Δ E = W \neq 0$

VITEEE-2019

Thermodynamics

272400 The values of $Δ H$ and $Δ S$ for the reaction, $C$ (graphite) $+ {CO}_{2} (g) \to 2 CO (g)$ are $170 kJ$ and $170 {JK}^{1}$ , respectively. This reaction will be spontaneous at

1

910 K

2

1000 K

3

510 K

4

710 K

Explanation:

Given-
$Δ H = 170 kJ$
$Δ S = 170 {JK}^{- 1}$
$T = ?$
Now, $Δ G = Δ H - T Δ S$ (at equilibrium $Δ G = 0$ )
$∴ Δ H = T Δ S$ or $T = \frac{Δ H}{Δ S}, T = \frac{170}{170 \times 10^{- 3}}$
$T = 1000 K$
Therefore, this reaction is spontaneous at this temperature.

AIPMT-2009

Thermodynamics

272401 Calculate enthalpy change for the change $8 S (g) \to S_{8} (g)$ , given that
$H_{2} S_{2} (g) \to 2 H (g) + 2 S (g), Δ H = 239.0 k cal {mol}^{- 1}$
$H_{2} S_{2} (g) \to 2 H (g) + S (g), Δ H = 175.0 k cal {mol}^{- 1}$

1

+ 512.0 k cal

2

- 512.0 k cal

3

508.8 k cal

4

- 508.0 k cal

Explanation:

Given that-
$H_{2} S_{2} (g) \to 2 H (g) + 2 S (g) Δ H = 239.0 k cal {mol}^{- 1}$
$H_{2} S_{2} (g) \to 2 H (g) + S (g) Δ H = 175.0 k cal {mol}^{- 1}$
$Δ H_{S - S} + 2 Δ H_{H - S} = 239$
And $2 Δ H_{H - S} = 175$
Hence, $2 Δ H_{S - S} = 239 - 175 = 64 k cal {mol}^{- 1}$
Then, $Δ H$ for $8 S (g) \to S_{S} (g)$ is
$8 \times (- 64) = - 512 k cal$

VITEEE-2018

Thermodynamics

272403 Consider the reaction: $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ carried out at constant temperature and pressure. If $Δ H$ and $Δ U$ are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

1

Δ H > Δ U

2

Δ H < Δ U

3

Δ H = Δ U

4

Δ H = 0

Explanation:

Given the reaction -
$N_{2} + 3 H_{2} \to 2 {NH}_{3}$
$Δ n = 2 - (3 + 1) = - 2$
$∴ Δ H = Δ U + Δ nRT$
$Δ H = Δ U - 2 RT$
So, $Δ U > Δ H$

AIEEE-2005

Thermodynamics

272398 For the reaction
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), Δ H$
$= - 285.8 kJ {mol}^{- 1}$
$Δ S = - 0.163 kJ {mol}^{- 1} K^{- 1}$
What is the value of free energy change at $27^{\circ} C$ for the reaction?

1

- 236.9 kJ {mol}^{- 1}

2

- 281.4 kJ {mol}^{- 1}

3

- 334.7 kJ {mol}^{- 1}

4

+ 334.7 kJ {mol}^{- 1}

Explanation:

Given,
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$
$Δ H = - 285.8 kJ / mole$
$Δ S = - 0.163 kJ / mole$
$Δ G = Δ H - T Δ S$
$Δ G = - 285.8 + 300 \times 0.163$
$Δ G = - 236.97 kJ / mole$

Thermodynamics

272399 In a closed insulated container, a liquid is stirred with a paddle to increase its temperature. In this process, which of the following is true?

1

Δ E = W = Q = 0

2

Δ E \neq 0, Q = W = 0

3

Δ E = W \neq 0, Q = 0

4

Δ E = Q \neq 0, W = 0

Explanation:

According to $I I^{st}$ Law of Thermodynamic
$q = Δ E + W$
$E =$ change in internal energy
$W =$ work done on system
Since system is closed and insulated, $q = 0$
Paddle work is done on system.
Hence internal energy of the system is increases.
$∴ Δ E = W \neq 0$

VITEEE-2019

Thermodynamics

272400 The values of $Δ H$ and $Δ S$ for the reaction, $C$ (graphite) $+ {CO}_{2} (g) \to 2 CO (g)$ are $170 kJ$ and $170 {JK}^{1}$ , respectively. This reaction will be spontaneous at

1

910 K

2

1000 K

3

510 K

4

710 K

Explanation:

Given-
$Δ H = 170 kJ$
$Δ S = 170 {JK}^{- 1}$
$T = ?$
Now, $Δ G = Δ H - T Δ S$ (at equilibrium $Δ G = 0$ )
$∴ Δ H = T Δ S$ or $T = \frac{Δ H}{Δ S}, T = \frac{170}{170 \times 10^{- 3}}$
$T = 1000 K$
Therefore, this reaction is spontaneous at this temperature.

AIPMT-2009

Thermodynamics

272401 Calculate enthalpy change for the change $8 S (g) \to S_{8} (g)$ , given that
$H_{2} S_{2} (g) \to 2 H (g) + 2 S (g), Δ H = 239.0 k cal {mol}^{- 1}$
$H_{2} S_{2} (g) \to 2 H (g) + S (g), Δ H = 175.0 k cal {mol}^{- 1}$

1

+ 512.0 k cal

2

- 512.0 k cal

3

508.8 k cal

4

- 508.0 k cal

Explanation:

Given that-
$H_{2} S_{2} (g) \to 2 H (g) + 2 S (g) Δ H = 239.0 k cal {mol}^{- 1}$
$H_{2} S_{2} (g) \to 2 H (g) + S (g) Δ H = 175.0 k cal {mol}^{- 1}$
$Δ H_{S - S} + 2 Δ H_{H - S} = 239$
And $2 Δ H_{H - S} = 175$
Hence, $2 Δ H_{S - S} = 239 - 175 = 64 k cal {mol}^{- 1}$
Then, $Δ H$ for $8 S (g) \to S_{S} (g)$ is
$8 \times (- 64) = - 512 k cal$

VITEEE-2018

Thermodynamics

272403 Consider the reaction: $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ carried out at constant temperature and pressure. If $Δ H$ and $Δ U$ are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

1

Δ H > Δ U

2

Δ H < Δ U

3

Δ H = Δ U

4

Δ H = 0

Explanation:

Given the reaction -
$N_{2} + 3 H_{2} \to 2 {NH}_{3}$
$Δ n = 2 - (3 + 1) = - 2$
$∴ Δ H = Δ U + Δ nRT$
$Δ H = Δ U - 2 RT$
So, $Δ U > Δ H$

AIEEE-2005

Thermodynamics

272398 For the reaction
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), Δ H$
$= - 285.8 kJ {mol}^{- 1}$
$Δ S = - 0.163 kJ {mol}^{- 1} K^{- 1}$
What is the value of free energy change at $27^{\circ} C$ for the reaction?

1

- 236.9 kJ {mol}^{- 1}

2

- 281.4 kJ {mol}^{- 1}

3

- 334.7 kJ {mol}^{- 1}

4

+ 334.7 kJ {mol}^{- 1}

Explanation:

Given,
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$
$Δ H = - 285.8 kJ / mole$
$Δ S = - 0.163 kJ / mole$
$Δ G = Δ H - T Δ S$
$Δ G = - 285.8 + 300 \times 0.163$
$Δ G = - 236.97 kJ / mole$

Thermodynamics

272399 In a closed insulated container, a liquid is stirred with a paddle to increase its temperature. In this process, which of the following is true?

1

Δ E = W = Q = 0

2

Δ E \neq 0, Q = W = 0

3

Δ E = W \neq 0, Q = 0

4

Δ E = Q \neq 0, W = 0

Explanation:

According to $I I^{st}$ Law of Thermodynamic
$q = Δ E + W$
$E =$ change in internal energy
$W =$ work done on system
Since system is closed and insulated, $q = 0$
Paddle work is done on system.
Hence internal energy of the system is increases.
$∴ Δ E = W \neq 0$

VITEEE-2019

Thermodynamics

272400 The values of $Δ H$ and $Δ S$ for the reaction, $C$ (graphite) $+ {CO}_{2} (g) \to 2 CO (g)$ are $170 kJ$ and $170 {JK}^{1}$ , respectively. This reaction will be spontaneous at

1

910 K

2

1000 K

3

510 K

4

710 K

Explanation:

Given-
$Δ H = 170 kJ$
$Δ S = 170 {JK}^{- 1}$
$T = ?$
Now, $Δ G = Δ H - T Δ S$ (at equilibrium $Δ G = 0$ )
$∴ Δ H = T Δ S$ or $T = \frac{Δ H}{Δ S}, T = \frac{170}{170 \times 10^{- 3}}$
$T = 1000 K$
Therefore, this reaction is spontaneous at this temperature.

AIPMT-2009

Thermodynamics

272401 Calculate enthalpy change for the change $8 S (g) \to S_{8} (g)$ , given that
$H_{2} S_{2} (g) \to 2 H (g) + 2 S (g), Δ H = 239.0 k cal {mol}^{- 1}$
$H_{2} S_{2} (g) \to 2 H (g) + S (g), Δ H = 175.0 k cal {mol}^{- 1}$

1

+ 512.0 k cal

2

- 512.0 k cal

3

508.8 k cal

4

- 508.0 k cal

Explanation:

Given that-
$H_{2} S_{2} (g) \to 2 H (g) + 2 S (g) Δ H = 239.0 k cal {mol}^{- 1}$
$H_{2} S_{2} (g) \to 2 H (g) + S (g) Δ H = 175.0 k cal {mol}^{- 1}$
$Δ H_{S - S} + 2 Δ H_{H - S} = 239$
And $2 Δ H_{H - S} = 175$
Hence, $2 Δ H_{S - S} = 239 - 175 = 64 k cal {mol}^{- 1}$
Then, $Δ H$ for $8 S (g) \to S_{S} (g)$ is
$8 \times (- 64) = - 512 k cal$

VITEEE-2018

Thermodynamics

272403 Consider the reaction: $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ carried out at constant temperature and pressure. If $Δ H$ and $Δ U$ are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

1

Δ H > Δ U

2

Δ H < Δ U

3

Δ H = Δ U

4

Δ H = 0

Explanation:

Given the reaction -
$N_{2} + 3 H_{2} \to 2 {NH}_{3}$
$Δ n = 2 - (3 + 1) = - 2$
$∴ Δ H = Δ U + Δ nRT$
$Δ H = Δ U - 2 RT$
So, $Δ U > Δ H$

AIEEE-2005

Thermodynamics

272398 For the reaction
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), Δ H$
$= - 285.8 kJ {mol}^{- 1}$
$Δ S = - 0.163 kJ {mol}^{- 1} K^{- 1}$
What is the value of free energy change at $27^{\circ} C$ for the reaction?

1

- 236.9 kJ {mol}^{- 1}

2

- 281.4 kJ {mol}^{- 1}

3

- 334.7 kJ {mol}^{- 1}

4

+ 334.7 kJ {mol}^{- 1}

Explanation:

Given,
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$
$Δ H = - 285.8 kJ / mole$
$Δ S = - 0.163 kJ / mole$
$Δ G = Δ H - T Δ S$
$Δ G = - 285.8 + 300 \times 0.163$
$Δ G = - 236.97 kJ / mole$

Thermodynamics

272399 In a closed insulated container, a liquid is stirred with a paddle to increase its temperature. In this process, which of the following is true?

1

Δ E = W = Q = 0

2

Δ E \neq 0, Q = W = 0

3

Δ E = W \neq 0, Q = 0

4

Δ E = Q \neq 0, W = 0

Explanation:

According to $I I^{st}$ Law of Thermodynamic
$q = Δ E + W$
$E =$ change in internal energy
$W =$ work done on system
Since system is closed and insulated, $q = 0$
Paddle work is done on system.
Hence internal energy of the system is increases.
$∴ Δ E = W \neq 0$

VITEEE-2019

Thermodynamics

272400 The values of $Δ H$ and $Δ S$ for the reaction, $C$ (graphite) $+ {CO}_{2} (g) \to 2 CO (g)$ are $170 kJ$ and $170 {JK}^{1}$ , respectively. This reaction will be spontaneous at

1

910 K

2

1000 K

3

510 K

4

710 K

Explanation:

Given-
$Δ H = 170 kJ$
$Δ S = 170 {JK}^{- 1}$
$T = ?$
Now, $Δ G = Δ H - T Δ S$ (at equilibrium $Δ G = 0$ )
$∴ Δ H = T Δ S$ or $T = \frac{Δ H}{Δ S}, T = \frac{170}{170 \times 10^{- 3}}$
$T = 1000 K$
Therefore, this reaction is spontaneous at this temperature.

AIPMT-2009

Thermodynamics

272401 Calculate enthalpy change for the change $8 S (g) \to S_{8} (g)$ , given that
$H_{2} S_{2} (g) \to 2 H (g) + 2 S (g), Δ H = 239.0 k cal {mol}^{- 1}$
$H_{2} S_{2} (g) \to 2 H (g) + S (g), Δ H = 175.0 k cal {mol}^{- 1}$

1

+ 512.0 k cal

2

- 512.0 k cal

3

508.8 k cal

4

- 508.0 k cal

Explanation:

Given that-
$H_{2} S_{2} (g) \to 2 H (g) + 2 S (g) Δ H = 239.0 k cal {mol}^{- 1}$
$H_{2} S_{2} (g) \to 2 H (g) + S (g) Δ H = 175.0 k cal {mol}^{- 1}$
$Δ H_{S - S} + 2 Δ H_{H - S} = 239$
And $2 Δ H_{H - S} = 175$
Hence, $2 Δ H_{S - S} = 239 - 175 = 64 k cal {mol}^{- 1}$
Then, $Δ H$ for $8 S (g) \to S_{S} (g)$ is
$8 \times (- 64) = - 512 k cal$

VITEEE-2018

Thermodynamics

272403 Consider the reaction: $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ carried out at constant temperature and pressure. If $Δ H$ and $Δ U$ are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

1

Δ H > Δ U

2

Δ H < Δ U

3

Δ H = Δ U

4

Δ H = 0

Explanation:

Given the reaction -
$N_{2} + 3 H_{2} \to 2 {NH}_{3}$
$Δ n = 2 - (3 + 1) = - 2$
$∴ Δ H = Δ U + Δ nRT$
$Δ H = Δ U - 2 RT$
So, $Δ U > Δ H$

AIEEE-2005

Thermodynamics

272398 For the reaction
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l), Δ H$
$= - 285.8 kJ {mol}^{- 1}$
$Δ S = - 0.163 kJ {mol}^{- 1} K^{- 1}$
What is the value of free energy change at $27^{\circ} C$ for the reaction?

1

- 236.9 kJ {mol}^{- 1}

2

- 281.4 kJ {mol}^{- 1}

3

- 334.7 kJ {mol}^{- 1}

4

+ 334.7 kJ {mol}^{- 1}

Explanation:

Given,
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$
$Δ H = - 285.8 kJ / mole$
$Δ S = - 0.163 kJ / mole$
$Δ G = Δ H - T Δ S$
$Δ G = - 285.8 + 300 \times 0.163$
$Δ G = - 236.97 kJ / mole$

Thermodynamics

272399 In a closed insulated container, a liquid is stirred with a paddle to increase its temperature. In this process, which of the following is true?

1

Δ E = W = Q = 0

2

Δ E \neq 0, Q = W = 0

3

Δ E = W \neq 0, Q = 0

4

Δ E = Q \neq 0, W = 0

Explanation:

According to $I I^{st}$ Law of Thermodynamic
$q = Δ E + W$
$E =$ change in internal energy
$W =$ work done on system
Since system is closed and insulated, $q = 0$
Paddle work is done on system.
Hence internal energy of the system is increases.
$∴ Δ E = W \neq 0$

VITEEE-2019

Thermodynamics

272400 The values of $Δ H$ and $Δ S$ for the reaction, $C$ (graphite) $+ {CO}_{2} (g) \to 2 CO (g)$ are $170 kJ$ and $170 {JK}^{1}$ , respectively. This reaction will be spontaneous at

1

910 K

2

1000 K

3

510 K

4

710 K

Explanation:

Given-
$Δ H = 170 kJ$
$Δ S = 170 {JK}^{- 1}$
$T = ?$
Now, $Δ G = Δ H - T Δ S$ (at equilibrium $Δ G = 0$ )
$∴ Δ H = T Δ S$ or $T = \frac{Δ H}{Δ S}, T = \frac{170}{170 \times 10^{- 3}}$
$T = 1000 K$
Therefore, this reaction is spontaneous at this temperature.

AIPMT-2009

Thermodynamics

272401 Calculate enthalpy change for the change $8 S (g) \to S_{8} (g)$ , given that
$H_{2} S_{2} (g) \to 2 H (g) + 2 S (g), Δ H = 239.0 k cal {mol}^{- 1}$
$H_{2} S_{2} (g) \to 2 H (g) + S (g), Δ H = 175.0 k cal {mol}^{- 1}$

1

+ 512.0 k cal

2

- 512.0 k cal

3

508.8 k cal

4

- 508.0 k cal

Explanation:

Given that-
$H_{2} S_{2} (g) \to 2 H (g) + 2 S (g) Δ H = 239.0 k cal {mol}^{- 1}$
$H_{2} S_{2} (g) \to 2 H (g) + S (g) Δ H = 175.0 k cal {mol}^{- 1}$
$Δ H_{S - S} + 2 Δ H_{H - S} = 239$
And $2 Δ H_{H - S} = 175$
Hence, $2 Δ H_{S - S} = 239 - 175 = 64 k cal {mol}^{- 1}$
Then, $Δ H$ for $8 S (g) \to S_{S} (g)$ is
$8 \times (- 64) = - 512 k cal$

VITEEE-2018

Thermodynamics

272403 Consider the reaction: $N_{2} + 3 H_{2} \to 2 {NH}_{3}$ carried out at constant temperature and pressure. If $Δ H$ and $Δ U$ are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

1

Δ H > Δ U

2

Δ H < Δ U

3

Δ H = Δ U

4

Δ H = 0

Explanation:

Given the reaction -
$N_{2} + 3 H_{2} \to 2 {NH}_{3}$
$Δ n = 2 - (3 + 1) = - 2$
$∴ Δ H = Δ U + Δ nRT$
$Δ H = Δ U - 2 RT$
So, $Δ U > Δ H$

AIEEE-2005

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