QBManager

Thermodynamics

273073 Which of the following relation is correct?

1

Δ H = Δ G + T Δ S

2

Δ H = - Δ G - T Δ S

3

Δ H = Δ G - T Δ S

4

Δ H = - Δ G + T Δ S

Explanation:

Gibbs Free Energy
$Δ H = Δ G + T Δ S$
Where, $Δ H =$ enthalpy change
$\begin{aligned} Δ S & = entropy change \\ T & = temperature \\ Δ H & = Δ G + T Δ S \\ Δ G & = Δ H - T Δ S \end{aligned}$

CG PET - 2004

Thermodynamics

273075 In an irreverible process taking place at constant $T$ and $p$ and in which only pressure-
volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria

1

(dS)_{V, E} < 0, (dG)_{T, p} < 0

2

(dS)_{V, E} > 0, (dG)_{T, p} < 0

3

(dS)_{V, E} = 0, (dG)_{T, p} = 0

4

(dS)_{V, E} = 0, (dG)_{T, p} > 0

Explanation:

An irreversible process (i.e. spontaneous
process )
$(ds)_{V, E}$ (change in entropy) $= + ve > 0$
$(dG)_{T, p}$ (change in Gibbs free energy) $= - ve < 0$

AIEEE-2003

Thermodynamics

273078 The incorrect expression among the following is

1

\frac{Δ G_{system}}{Δ G_{total}} = - T

2 In isothermal process,

W_{reversible} = - nRT In \frac{V_{f}}{V_{i}}

3

\ln K = \frac{Δ H^{\circ} - T Δ S^{\circ}}{RT}

4

K = e^{- Δ G^{\circ} / RT}

Explanation:

$Δ G^{\circ} = - RT \ln K$
(i)
$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$
from equation (i) and (ii)
$\begin{aligned} - (Δ H^{\circ} + T Δ S^{\circ}) = RT \ln K \\ \ln K = \frac{Δ H^{\circ} - T Δ S^{\circ}}{RT} \end{aligned}$

AIEEE-2012

Thermodynamics

273084 When compared to $Δ G^{\circ}$ for the formation of ${Al}_{2} O_{3}$ , the $Δ G^{0}$ for the formation of ${Cr}_{2} O_{3}$ is

1 Higher

2 Lower

3 Unpredicted

4 Same

Explanation:

Between ${Al}_{2} O_{3}$ and ${Cr}_{2} O_{3}, {Al}_{2} O_{3}$ is more stable, so its ease of formation is high i.e., $Δ G^{\circ}$ is negative or less than that of ${Cr}_{2} O_{3}$ .
Hence, $Δ G^{\circ}$ of ${Cr}_{2} O_{3} > Δ G^{\circ}$ of ${Al}_{2} O_{3}$ .

Karnataka-CET, 2008

Thermodynamics

273073 Which of the following relation is correct?

1

Δ H = Δ G + T Δ S

2

Δ H = - Δ G - T Δ S

3

Δ H = Δ G - T Δ S

4

Δ H = - Δ G + T Δ S

Explanation:

Gibbs Free Energy
$Δ H = Δ G + T Δ S$
Where, $Δ H =$ enthalpy change
$\begin{aligned} Δ S & = entropy change \\ T & = temperature \\ Δ H & = Δ G + T Δ S \\ Δ G & = Δ H - T Δ S \end{aligned}$

CG PET - 2004

Thermodynamics

273075 In an irreverible process taking place at constant $T$ and $p$ and in which only pressure-
volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria

1

(dS)_{V, E} < 0, (dG)_{T, p} < 0

2

(dS)_{V, E} > 0, (dG)_{T, p} < 0

3

(dS)_{V, E} = 0, (dG)_{T, p} = 0

4

(dS)_{V, E} = 0, (dG)_{T, p} > 0

Explanation:

An irreversible process (i.e. spontaneous
process )
$(ds)_{V, E}$ (change in entropy) $= + ve > 0$
$(dG)_{T, p}$ (change in Gibbs free energy) $= - ve < 0$

AIEEE-2003

Thermodynamics

273078 The incorrect expression among the following is

1

\frac{Δ G_{system}}{Δ G_{total}} = - T

2 In isothermal process,

W_{reversible} = - nRT In \frac{V_{f}}{V_{i}}

3

\ln K = \frac{Δ H^{\circ} - T Δ S^{\circ}}{RT}

4

K = e^{- Δ G^{\circ} / RT}

Explanation:

$Δ G^{\circ} = - RT \ln K$
(i)
$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$
from equation (i) and (ii)
$\begin{aligned} - (Δ H^{\circ} + T Δ S^{\circ}) = RT \ln K \\ \ln K = \frac{Δ H^{\circ} - T Δ S^{\circ}}{RT} \end{aligned}$

AIEEE-2012

Thermodynamics

273084 When compared to $Δ G^{\circ}$ for the formation of ${Al}_{2} O_{3}$ , the $Δ G^{0}$ for the formation of ${Cr}_{2} O_{3}$ is

1 Higher

2 Lower

3 Unpredicted

4 Same

Explanation:

Between ${Al}_{2} O_{3}$ and ${Cr}_{2} O_{3}, {Al}_{2} O_{3}$ is more stable, so its ease of formation is high i.e., $Δ G^{\circ}$ is negative or less than that of ${Cr}_{2} O_{3}$ .
Hence, $Δ G^{\circ}$ of ${Cr}_{2} O_{3} > Δ G^{\circ}$ of ${Al}_{2} O_{3}$ .

Karnataka-CET, 2008

Thermodynamics

273073 Which of the following relation is correct?

1

Δ H = Δ G + T Δ S

2

Δ H = - Δ G - T Δ S

3

Δ H = Δ G - T Δ S

4

Δ H = - Δ G + T Δ S

Explanation:

Gibbs Free Energy
$Δ H = Δ G + T Δ S$
Where, $Δ H =$ enthalpy change
$\begin{aligned} Δ S & = entropy change \\ T & = temperature \\ Δ H & = Δ G + T Δ S \\ Δ G & = Δ H - T Δ S \end{aligned}$

CG PET - 2004

Thermodynamics

273075 In an irreverible process taking place at constant $T$ and $p$ and in which only pressure-
volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria

1

(dS)_{V, E} < 0, (dG)_{T, p} < 0

2

(dS)_{V, E} > 0, (dG)_{T, p} < 0

3

(dS)_{V, E} = 0, (dG)_{T, p} = 0

4

(dS)_{V, E} = 0, (dG)_{T, p} > 0

Explanation:

An irreversible process (i.e. spontaneous
process )
$(ds)_{V, E}$ (change in entropy) $= + ve > 0$
$(dG)_{T, p}$ (change in Gibbs free energy) $= - ve < 0$

AIEEE-2003

Thermodynamics

273078 The incorrect expression among the following is

1

\frac{Δ G_{system}}{Δ G_{total}} = - T

2 In isothermal process,

W_{reversible} = - nRT In \frac{V_{f}}{V_{i}}

3

\ln K = \frac{Δ H^{\circ} - T Δ S^{\circ}}{RT}

4

K = e^{- Δ G^{\circ} / RT}

Explanation:

$Δ G^{\circ} = - RT \ln K$
(i)
$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$
from equation (i) and (ii)
$\begin{aligned} - (Δ H^{\circ} + T Δ S^{\circ}) = RT \ln K \\ \ln K = \frac{Δ H^{\circ} - T Δ S^{\circ}}{RT} \end{aligned}$

AIEEE-2012

Thermodynamics

273084 When compared to $Δ G^{\circ}$ for the formation of ${Al}_{2} O_{3}$ , the $Δ G^{0}$ for the formation of ${Cr}_{2} O_{3}$ is

1 Higher

2 Lower

3 Unpredicted

4 Same

Explanation:

Between ${Al}_{2} O_{3}$ and ${Cr}_{2} O_{3}, {Al}_{2} O_{3}$ is more stable, so its ease of formation is high i.e., $Δ G^{\circ}$ is negative or less than that of ${Cr}_{2} O_{3}$ .
Hence, $Δ G^{\circ}$ of ${Cr}_{2} O_{3} > Δ G^{\circ}$ of ${Al}_{2} O_{3}$ .

Karnataka-CET, 2008

Thermodynamics

273073 Which of the following relation is correct?

1

Δ H = Δ G + T Δ S

2

Δ H = - Δ G - T Δ S

3

Δ H = Δ G - T Δ S

4

Δ H = - Δ G + T Δ S

Explanation:

Gibbs Free Energy
$Δ H = Δ G + T Δ S$
Where, $Δ H =$ enthalpy change
$\begin{aligned} Δ S & = entropy change \\ T & = temperature \\ Δ H & = Δ G + T Δ S \\ Δ G & = Δ H - T Δ S \end{aligned}$

CG PET - 2004

Thermodynamics

273075 In an irreverible process taking place at constant $T$ and $p$ and in which only pressure-
volume work is being done, the change in Gibbs free energy (dG) and change in entropy (dS), satisfy the criteria

1

(dS)_{V, E} < 0, (dG)_{T, p} < 0

2

(dS)_{V, E} > 0, (dG)_{T, p} < 0

3

(dS)_{V, E} = 0, (dG)_{T, p} = 0

4

(dS)_{V, E} = 0, (dG)_{T, p} > 0

Explanation:

An irreversible process (i.e. spontaneous
process )
$(ds)_{V, E}$ (change in entropy) $= + ve > 0$
$(dG)_{T, p}$ (change in Gibbs free energy) $= - ve < 0$

AIEEE-2003

Thermodynamics

273078 The incorrect expression among the following is

1

\frac{Δ G_{system}}{Δ G_{total}} = - T

2 In isothermal process,

W_{reversible} = - nRT In \frac{V_{f}}{V_{i}}

3

\ln K = \frac{Δ H^{\circ} - T Δ S^{\circ}}{RT}

4

K = e^{- Δ G^{\circ} / RT}

Explanation:

$Δ G^{\circ} = - RT \ln K$
(i)
$Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$
from equation (i) and (ii)
$\begin{aligned} - (Δ H^{\circ} + T Δ S^{\circ}) = RT \ln K \\ \ln K = \frac{Δ H^{\circ} - T Δ S^{\circ}}{RT} \end{aligned}$

AIEEE-2012

Thermodynamics

273084 When compared to $Δ G^{\circ}$ for the formation of ${Al}_{2} O_{3}$ , the $Δ G^{0}$ for the formation of ${Cr}_{2} O_{3}$ is

1 Higher

2 Lower

3 Unpredicted

4 Same

Explanation:

Between ${Al}_{2} O_{3}$ and ${Cr}_{2} O_{3}, {Al}_{2} O_{3}$ is more stable, so its ease of formation is high i.e., $Δ G^{\circ}$ is negative or less than that of ${Cr}_{2} O_{3}$ .
Hence, $Δ G^{\circ}$ of ${Cr}_{2} O_{3} > Δ G^{\circ}$ of ${Al}_{2} O_{3}$ .

Karnataka-CET, 2008