QBManager

Thermodynamics

273082 The incorrect match in the following is

1

Δ G^{\circ} < 0, K > 1

2

Δ G^{\circ} = 0, K = 1

3

Δ G^{\circ} > 0, K < 1

4

Δ G^{\circ} < 0, K < 1

Explanation:

$Δ G = Δ G^{\circ} + RT \ln K$
At equilibrium $(Δ G = 0)$
$\begin{aligned} Δ G^{\circ} + RT \ln K = 0 \\ Δ G^{\circ} = - RT \ln K \end{aligned}$
If $K < 1$
Then $Δ G^{\circ} > 0$

JEE Main-2019, Shift-II

Thermodynamics

273053 Standard entropies of $X_{2}, Y_{2}$ and $X Y_{3}$ are 60 , 40 and $50 / K /$ mol respectively. At what temperature the following reaction will be at equilibrium.
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2}$ ? ${X Y}_{3}, Δ H = - 30 kJ$

1

500 K

2

750 K

3

1000 K

4

1250 K

Explanation:

Given, Standard entropies,
$\begin{aligned} S_{X_{2}} & = 60 {Jk}^{- 1} {mol}^{- 1} and Δ H = - 30 kJ \\ S_{Y_{2}} & = 40 {Jk}^{- 1} {mol}^{- 1} \\ S_{{XY}_{3}} & = 50 {Jk}^{- 1} {mol}^{- 1} \\ \frac{1}{2} X_{2} & + \frac{3}{2} Y_{2} ◻ XY \\ Δ S & = \sum S_{(Product)} - \sum S_{(Reactant)} \\ ∴ Δ S_{Reaction} & = 50 - \frac{3}{2} \times 40 - \frac{1}{2} \times 60 = 50 - 60 - 30 \end{aligned}$
$= - 40 {Jk}^{- 1} {mol}^{- 1}$
From the Gibb's free energy equation-
$Δ G = Δ H - T Δ S$ (for equilibrium $Δ G = 0$ )
$∴ Δ H = T Δ S$
Or $T = \frac{Δ H}{Δ S} = \frac{- 30 \times 1000}{- 40} = 750 K$

AP EAPCET 19-08-2021 Shift-I

Thermodynamics

273054 The temperature of $K$ at which $Δ G = 0$ , for a given reaction with $Δ H = - 20.5 kJ {mol}^{- 1}$ and $Δ S$ $= - 50.0 {JK}^{- 1} {mol}^{- 1}$ is

1 -410

2 410

3 2.44

4 0.36

Explanation:

Given, $Δ H = - 20.5 kJ {mol}^{- 1}$
$\begin{aligned} Δ S = - 50 {Jk}^{- 1} {mol}^{- 1} \\ Δ G = Δ H - T Δ S \\ 0 = - (20500) - T (- 50) (Δ G = 0) \\ 20500 = 50 T \\ T = 410 K \end{aligned}$

AP EAMCET (Engg.) -2014

Thermodynamics

273056 The standard enthalpy of the decomposition of $N_{2} O_{4}$ to ${NO}_{2}$ is $58.04 kJ$ and standard entropy of this reaction is $176.7 J / K$ . Therefore the standard free energy changes for this reaction at $25^{\circ} C$ is

1

- 5.39 kJ

2

539 kJ

3

5.39 kJ

4

- 539 kJ

Explanation:

Given, that
Standard enthalpy of decompositions $(Δ H) = 58.04 kJ$
Standard entropy $(Δ S) = 176.7 J / K$
$= 176.7 \times 10^{- 3} kJ / K$
And temperature $(T) = 25^{\circ} C + 273 = 298 K$ .
Change in free energy $(Δ G) = Δ H - T Δ S$
$\begin{aligned} = 58.04 - 298 \times (176.7 \times 10^{- 3}) = 58.04 - 52.65 \\ = 5.39 kJ \end{aligned}$

COMEDK-2017

Thermodynamics

273057 The equilibrium constant of a reaction is 0.008 at $298 K$ . The standard free energy change of the reaction at the same temperature is

1

+ 11.96 kJ

2

- 11.96 kJ

3

- 5.43 kJ

4

- 8.46 kJ

Explanation:

Given, Equilibrium constant $(K) = 0.008$ Temperature $(T) = 298 K, R = 8.314 J {mol}^{- 1} K^{- 1}$ $Δ G^{\circ} = - 2.303 RT \log K$
$\begin{aligned} Δ G^{\circ} = - 2.303 \times 8.314 \times 298 \times \log (0.008) \\ Δ G^{\circ} = - 5705.84 \times - 2.09 \\ Δ G^{\circ} = 11964.65 = 11.96 kJ \end{aligned}$

COMEDK-2020

Thermodynamics

273082 The incorrect match in the following is

1

Δ G^{\circ} < 0, K > 1

2

Δ G^{\circ} = 0, K = 1

3

Δ G^{\circ} > 0, K < 1

4

Δ G^{\circ} < 0, K < 1

Explanation:

$Δ G = Δ G^{\circ} + RT \ln K$
At equilibrium $(Δ G = 0)$
$\begin{aligned} Δ G^{\circ} + RT \ln K = 0 \\ Δ G^{\circ} = - RT \ln K \end{aligned}$
If $K < 1$
Then $Δ G^{\circ} > 0$

JEE Main-2019, Shift-II

Thermodynamics

273053 Standard entropies of $X_{2}, Y_{2}$ and $X Y_{3}$ are 60 , 40 and $50 / K /$ mol respectively. At what temperature the following reaction will be at equilibrium.
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2}$ ? ${X Y}_{3}, Δ H = - 30 kJ$

1

500 K

2

750 K

3

1000 K

4

1250 K

Explanation:

Given, Standard entropies,
$\begin{aligned} S_{X_{2}} & = 60 {Jk}^{- 1} {mol}^{- 1} and Δ H = - 30 kJ \\ S_{Y_{2}} & = 40 {Jk}^{- 1} {mol}^{- 1} \\ S_{{XY}_{3}} & = 50 {Jk}^{- 1} {mol}^{- 1} \\ \frac{1}{2} X_{2} & + \frac{3}{2} Y_{2} ◻ XY \\ Δ S & = \sum S_{(Product)} - \sum S_{(Reactant)} \\ ∴ Δ S_{Reaction} & = 50 - \frac{3}{2} \times 40 - \frac{1}{2} \times 60 = 50 - 60 - 30 \end{aligned}$
$= - 40 {Jk}^{- 1} {mol}^{- 1}$
From the Gibb's free energy equation-
$Δ G = Δ H - T Δ S$ (for equilibrium $Δ G = 0$ )
$∴ Δ H = T Δ S$
Or $T = \frac{Δ H}{Δ S} = \frac{- 30 \times 1000}{- 40} = 750 K$

AP EAPCET 19-08-2021 Shift-I

Thermodynamics

273054 The temperature of $K$ at which $Δ G = 0$ , for a given reaction with $Δ H = - 20.5 kJ {mol}^{- 1}$ and $Δ S$ $= - 50.0 {JK}^{- 1} {mol}^{- 1}$ is

1 -410

2 410

3 2.44

4 0.36

Explanation:

Given, $Δ H = - 20.5 kJ {mol}^{- 1}$
$\begin{aligned} Δ S = - 50 {Jk}^{- 1} {mol}^{- 1} \\ Δ G = Δ H - T Δ S \\ 0 = - (20500) - T (- 50) (Δ G = 0) \\ 20500 = 50 T \\ T = 410 K \end{aligned}$

AP EAMCET (Engg.) -2014

Thermodynamics

273056 The standard enthalpy of the decomposition of $N_{2} O_{4}$ to ${NO}_{2}$ is $58.04 kJ$ and standard entropy of this reaction is $176.7 J / K$ . Therefore the standard free energy changes for this reaction at $25^{\circ} C$ is

1

- 5.39 kJ

2

539 kJ

3

5.39 kJ

4

- 539 kJ

Explanation:

Given, that
Standard enthalpy of decompositions $(Δ H) = 58.04 kJ$
Standard entropy $(Δ S) = 176.7 J / K$
$= 176.7 \times 10^{- 3} kJ / K$
And temperature $(T) = 25^{\circ} C + 273 = 298 K$ .
Change in free energy $(Δ G) = Δ H - T Δ S$
$\begin{aligned} = 58.04 - 298 \times (176.7 \times 10^{- 3}) = 58.04 - 52.65 \\ = 5.39 kJ \end{aligned}$

COMEDK-2017

Thermodynamics

273057 The equilibrium constant of a reaction is 0.008 at $298 K$ . The standard free energy change of the reaction at the same temperature is

1

+ 11.96 kJ

2

- 11.96 kJ

3

- 5.43 kJ

4

- 8.46 kJ

Explanation:

Given, Equilibrium constant $(K) = 0.008$ Temperature $(T) = 298 K, R = 8.314 J {mol}^{- 1} K^{- 1}$ $Δ G^{\circ} = - 2.303 RT \log K$
$\begin{aligned} Δ G^{\circ} = - 2.303 \times 8.314 \times 298 \times \log (0.008) \\ Δ G^{\circ} = - 5705.84 \times - 2.09 \\ Δ G^{\circ} = 11964.65 = 11.96 kJ \end{aligned}$

COMEDK-2020

Thermodynamics

273082 The incorrect match in the following is

1

Δ G^{\circ} < 0, K > 1

2

Δ G^{\circ} = 0, K = 1

3

Δ G^{\circ} > 0, K < 1

4

Δ G^{\circ} < 0, K < 1

Explanation:

$Δ G = Δ G^{\circ} + RT \ln K$
At equilibrium $(Δ G = 0)$
$\begin{aligned} Δ G^{\circ} + RT \ln K = 0 \\ Δ G^{\circ} = - RT \ln K \end{aligned}$
If $K < 1$
Then $Δ G^{\circ} > 0$

JEE Main-2019, Shift-II

Thermodynamics

273053 Standard entropies of $X_{2}, Y_{2}$ and $X Y_{3}$ are 60 , 40 and $50 / K /$ mol respectively. At what temperature the following reaction will be at equilibrium.
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2}$ ? ${X Y}_{3}, Δ H = - 30 kJ$

1

500 K

2

750 K

3

1000 K

4

1250 K

Explanation:

Given, Standard entropies,
$\begin{aligned} S_{X_{2}} & = 60 {Jk}^{- 1} {mol}^{- 1} and Δ H = - 30 kJ \\ S_{Y_{2}} & = 40 {Jk}^{- 1} {mol}^{- 1} \\ S_{{XY}_{3}} & = 50 {Jk}^{- 1} {mol}^{- 1} \\ \frac{1}{2} X_{2} & + \frac{3}{2} Y_{2} ◻ XY \\ Δ S & = \sum S_{(Product)} - \sum S_{(Reactant)} \\ ∴ Δ S_{Reaction} & = 50 - \frac{3}{2} \times 40 - \frac{1}{2} \times 60 = 50 - 60 - 30 \end{aligned}$
$= - 40 {Jk}^{- 1} {mol}^{- 1}$
From the Gibb's free energy equation-
$Δ G = Δ H - T Δ S$ (for equilibrium $Δ G = 0$ )
$∴ Δ H = T Δ S$
Or $T = \frac{Δ H}{Δ S} = \frac{- 30 \times 1000}{- 40} = 750 K$

AP EAPCET 19-08-2021 Shift-I

Thermodynamics

273054 The temperature of $K$ at which $Δ G = 0$ , for a given reaction with $Δ H = - 20.5 kJ {mol}^{- 1}$ and $Δ S$ $= - 50.0 {JK}^{- 1} {mol}^{- 1}$ is

1 -410

2 410

3 2.44

4 0.36

Explanation:

Given, $Δ H = - 20.5 kJ {mol}^{- 1}$
$\begin{aligned} Δ S = - 50 {Jk}^{- 1} {mol}^{- 1} \\ Δ G = Δ H - T Δ S \\ 0 = - (20500) - T (- 50) (Δ G = 0) \\ 20500 = 50 T \\ T = 410 K \end{aligned}$

AP EAMCET (Engg.) -2014

Thermodynamics

273056 The standard enthalpy of the decomposition of $N_{2} O_{4}$ to ${NO}_{2}$ is $58.04 kJ$ and standard entropy of this reaction is $176.7 J / K$ . Therefore the standard free energy changes for this reaction at $25^{\circ} C$ is

1

- 5.39 kJ

2

539 kJ

3

5.39 kJ

4

- 539 kJ

Explanation:

Given, that
Standard enthalpy of decompositions $(Δ H) = 58.04 kJ$
Standard entropy $(Δ S) = 176.7 J / K$
$= 176.7 \times 10^{- 3} kJ / K$
And temperature $(T) = 25^{\circ} C + 273 = 298 K$ .
Change in free energy $(Δ G) = Δ H - T Δ S$
$\begin{aligned} = 58.04 - 298 \times (176.7 \times 10^{- 3}) = 58.04 - 52.65 \\ = 5.39 kJ \end{aligned}$

COMEDK-2017

Thermodynamics

273057 The equilibrium constant of a reaction is 0.008 at $298 K$ . The standard free energy change of the reaction at the same temperature is

1

+ 11.96 kJ

2

- 11.96 kJ

3

- 5.43 kJ

4

- 8.46 kJ

Explanation:

Given, Equilibrium constant $(K) = 0.008$ Temperature $(T) = 298 K, R = 8.314 J {mol}^{- 1} K^{- 1}$ $Δ G^{\circ} = - 2.303 RT \log K$
$\begin{aligned} Δ G^{\circ} = - 2.303 \times 8.314 \times 298 \times \log (0.008) \\ Δ G^{\circ} = - 5705.84 \times - 2.09 \\ Δ G^{\circ} = 11964.65 = 11.96 kJ \end{aligned}$

COMEDK-2020

Thermodynamics

273082 The incorrect match in the following is

1

Δ G^{\circ} < 0, K > 1

2

Δ G^{\circ} = 0, K = 1

3

Δ G^{\circ} > 0, K < 1

4

Δ G^{\circ} < 0, K < 1

Explanation:

$Δ G = Δ G^{\circ} + RT \ln K$
At equilibrium $(Δ G = 0)$
$\begin{aligned} Δ G^{\circ} + RT \ln K = 0 \\ Δ G^{\circ} = - RT \ln K \end{aligned}$
If $K < 1$
Then $Δ G^{\circ} > 0$

JEE Main-2019, Shift-II

Thermodynamics

273053 Standard entropies of $X_{2}, Y_{2}$ and $X Y_{3}$ are 60 , 40 and $50 / K /$ mol respectively. At what temperature the following reaction will be at equilibrium.
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2}$ ? ${X Y}_{3}, Δ H = - 30 kJ$

1

500 K

2

750 K

3

1000 K

4

1250 K

Explanation:

Given, Standard entropies,
$\begin{aligned} S_{X_{2}} & = 60 {Jk}^{- 1} {mol}^{- 1} and Δ H = - 30 kJ \\ S_{Y_{2}} & = 40 {Jk}^{- 1} {mol}^{- 1} \\ S_{{XY}_{3}} & = 50 {Jk}^{- 1} {mol}^{- 1} \\ \frac{1}{2} X_{2} & + \frac{3}{2} Y_{2} ◻ XY \\ Δ S & = \sum S_{(Product)} - \sum S_{(Reactant)} \\ ∴ Δ S_{Reaction} & = 50 - \frac{3}{2} \times 40 - \frac{1}{2} \times 60 = 50 - 60 - 30 \end{aligned}$
$= - 40 {Jk}^{- 1} {mol}^{- 1}$
From the Gibb's free energy equation-
$Δ G = Δ H - T Δ S$ (for equilibrium $Δ G = 0$ )
$∴ Δ H = T Δ S$
Or $T = \frac{Δ H}{Δ S} = \frac{- 30 \times 1000}{- 40} = 750 K$

AP EAPCET 19-08-2021 Shift-I

Thermodynamics

273054 The temperature of $K$ at which $Δ G = 0$ , for a given reaction with $Δ H = - 20.5 kJ {mol}^{- 1}$ and $Δ S$ $= - 50.0 {JK}^{- 1} {mol}^{- 1}$ is

1 -410

2 410

3 2.44

4 0.36

Explanation:

Given, $Δ H = - 20.5 kJ {mol}^{- 1}$
$\begin{aligned} Δ S = - 50 {Jk}^{- 1} {mol}^{- 1} \\ Δ G = Δ H - T Δ S \\ 0 = - (20500) - T (- 50) (Δ G = 0) \\ 20500 = 50 T \\ T = 410 K \end{aligned}$

AP EAMCET (Engg.) -2014

Thermodynamics

273056 The standard enthalpy of the decomposition of $N_{2} O_{4}$ to ${NO}_{2}$ is $58.04 kJ$ and standard entropy of this reaction is $176.7 J / K$ . Therefore the standard free energy changes for this reaction at $25^{\circ} C$ is

1

- 5.39 kJ

2

539 kJ

3

5.39 kJ

4

- 539 kJ

Explanation:

Given, that
Standard enthalpy of decompositions $(Δ H) = 58.04 kJ$
Standard entropy $(Δ S) = 176.7 J / K$
$= 176.7 \times 10^{- 3} kJ / K$
And temperature $(T) = 25^{\circ} C + 273 = 298 K$ .
Change in free energy $(Δ G) = Δ H - T Δ S$
$\begin{aligned} = 58.04 - 298 \times (176.7 \times 10^{- 3}) = 58.04 - 52.65 \\ = 5.39 kJ \end{aligned}$

COMEDK-2017

Thermodynamics

273057 The equilibrium constant of a reaction is 0.008 at $298 K$ . The standard free energy change of the reaction at the same temperature is

1

+ 11.96 kJ

2

- 11.96 kJ

3

- 5.43 kJ

4

- 8.46 kJ

Explanation:

Given, Equilibrium constant $(K) = 0.008$ Temperature $(T) = 298 K, R = 8.314 J {mol}^{- 1} K^{- 1}$ $Δ G^{\circ} = - 2.303 RT \log K$
$\begin{aligned} Δ G^{\circ} = - 2.303 \times 8.314 \times 298 \times \log (0.008) \\ Δ G^{\circ} = - 5705.84 \times - 2.09 \\ Δ G^{\circ} = 11964.65 = 11.96 kJ \end{aligned}$

COMEDK-2020

Thermodynamics

273082 The incorrect match in the following is

1

Δ G^{\circ} < 0, K > 1

2

Δ G^{\circ} = 0, K = 1

3

Δ G^{\circ} > 0, K < 1

4

Δ G^{\circ} < 0, K < 1

Explanation:

$Δ G = Δ G^{\circ} + RT \ln K$
At equilibrium $(Δ G = 0)$
$\begin{aligned} Δ G^{\circ} + RT \ln K = 0 \\ Δ G^{\circ} = - RT \ln K \end{aligned}$
If $K < 1$
Then $Δ G^{\circ} > 0$

JEE Main-2019, Shift-II

Thermodynamics

273053 Standard entropies of $X_{2}, Y_{2}$ and $X Y_{3}$ are 60 , 40 and $50 / K /$ mol respectively. At what temperature the following reaction will be at equilibrium.
$\frac{1}{2} X_{2} + \frac{3}{2} Y_{2}$ ? ${X Y}_{3}, Δ H = - 30 kJ$

1

500 K

2

750 K

3

1000 K

4

1250 K

Explanation:

Given, Standard entropies,
$\begin{aligned} S_{X_{2}} & = 60 {Jk}^{- 1} {mol}^{- 1} and Δ H = - 30 kJ \\ S_{Y_{2}} & = 40 {Jk}^{- 1} {mol}^{- 1} \\ S_{{XY}_{3}} & = 50 {Jk}^{- 1} {mol}^{- 1} \\ \frac{1}{2} X_{2} & + \frac{3}{2} Y_{2} ◻ XY \\ Δ S & = \sum S_{(Product)} - \sum S_{(Reactant)} \\ ∴ Δ S_{Reaction} & = 50 - \frac{3}{2} \times 40 - \frac{1}{2} \times 60 = 50 - 60 - 30 \end{aligned}$
$= - 40 {Jk}^{- 1} {mol}^{- 1}$
From the Gibb's free energy equation-
$Δ G = Δ H - T Δ S$ (for equilibrium $Δ G = 0$ )
$∴ Δ H = T Δ S$
Or $T = \frac{Δ H}{Δ S} = \frac{- 30 \times 1000}{- 40} = 750 K$

AP EAPCET 19-08-2021 Shift-I

Thermodynamics

273054 The temperature of $K$ at which $Δ G = 0$ , for a given reaction with $Δ H = - 20.5 kJ {mol}^{- 1}$ and $Δ S$ $= - 50.0 {JK}^{- 1} {mol}^{- 1}$ is

1 -410

2 410

3 2.44

4 0.36

Explanation:

Given, $Δ H = - 20.5 kJ {mol}^{- 1}$
$\begin{aligned} Δ S = - 50 {Jk}^{- 1} {mol}^{- 1} \\ Δ G = Δ H - T Δ S \\ 0 = - (20500) - T (- 50) (Δ G = 0) \\ 20500 = 50 T \\ T = 410 K \end{aligned}$

AP EAMCET (Engg.) -2014

Thermodynamics

273056 The standard enthalpy of the decomposition of $N_{2} O_{4}$ to ${NO}_{2}$ is $58.04 kJ$ and standard entropy of this reaction is $176.7 J / K$ . Therefore the standard free energy changes for this reaction at $25^{\circ} C$ is

1

- 5.39 kJ

2

539 kJ

3

5.39 kJ

4

- 539 kJ

Explanation:

Given, that
Standard enthalpy of decompositions $(Δ H) = 58.04 kJ$
Standard entropy $(Δ S) = 176.7 J / K$
$= 176.7 \times 10^{- 3} kJ / K$
And temperature $(T) = 25^{\circ} C + 273 = 298 K$ .
Change in free energy $(Δ G) = Δ H - T Δ S$
$\begin{aligned} = 58.04 - 298 \times (176.7 \times 10^{- 3}) = 58.04 - 52.65 \\ = 5.39 kJ \end{aligned}$

COMEDK-2017

Thermodynamics

273057 The equilibrium constant of a reaction is 0.008 at $298 K$ . The standard free energy change of the reaction at the same temperature is

1

+ 11.96 kJ

2

- 11.96 kJ

3

- 5.43 kJ

4

- 8.46 kJ

Explanation:

Given, Equilibrium constant $(K) = 0.008$ Temperature $(T) = 298 K, R = 8.314 J {mol}^{- 1} K^{- 1}$ $Δ G^{\circ} = - 2.303 RT \log K$
$\begin{aligned} Δ G^{\circ} = - 2.303 \times 8.314 \times 298 \times \log (0.008) \\ Δ G^{\circ} = - 5705.84 \times - 2.09 \\ Δ G^{\circ} = 11964.65 = 11.96 kJ \end{aligned}$

COMEDK-2020