QBManager

Chemical Equilibrium

229333 Which of the following is a wrong statement about equilibrium state?

1 Rate of forward reaction= Rate of backward reaction

2 Equilibrium is dynamic

3 Catalysts increase value of equilibrium constant

4 Free energy change is zero

Explanation:

Chemical equilibrium is dynamic in nature which in other words means that reactions continue to occur in both forward as well as backward direction with the same speed.
Equilibrium constant has definite value for every chemical reaction at a given temperature. It is independent of concentration and catalyst.
Free energy change :
$Δ G = Δ G^{o} + 2.303 RT \log Q$
So, option (a), (b), (d) follows equilibrium state but (c) not follows.
Hence, option (c) is correct.

CG PET -2009

Chemical Equilibrium

229335 The standard emf of a cell, involving one electron change is found to be $0.591 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction is $(F =$ $96500 {Cmol}^{- 1}$ )

1

1.0 \times 10^{1}

2

1.0 \times 10^{5}

3

1.0 \times 10^{10}

4

1.0 \times 10^{30}

Explanation:

Relation between $K_{eq}$ and $E_{cell}$ us
$E_{cell}^{\circ} = \frac{2.303 RT}{nF} \log K_{eq}$
$E_{cell}^{\circ} = \frac{2.303 RT}{nF} \log K_{eq} ($ at $298 K)$
$E_{cell}^{\circ} = \frac{0.0591}{n} \log K_{eq}$
$0.0591 = \frac{0.0591}{n} \log K_{eq}$
$\log K_{eq} = 10 = K_{eq} = antilog 10$
$K_{eq} = 1 \times 10^{10}$

UPTU/UPSEE-2008

Chemical Equilibrium

229336 The equilibrium constant of the reaction
$Cu (s) + 2 {Ag}^{+} (aq) \to {Cu}^{2 +} (aq) + Ag (s)$
[Given : $E^{o} = 0.46 V$ ] will be

1

4 \times 10^{15}

2

4 \times 10^{14}

3

14 \times 10^{15}

4

15.6 \times 10^{5}

Explanation:

The equilibrium constant for the give reaction can be given as, [Given : $E^{o} = 0.46 V$ ]
$\begin{aligned} Cu (s) + 2 {Ag}^{+} (aq) \to {Cu}^{2 +} (aq) + Ag (s) \\ E^{\circ} = \frac{0.0591}{n} \log K \end{aligned}$
Where, $K =$ Equilibrium constant, $n =$ Number of electrons.
On putting the values in eqn, ....(i) we get
$0.46 = \frac{0.0591}{n} \log K$
$\log K = \frac{0.46 \times 2}{0.0591}$
$∴ K =$ Antilog 15.6
$= 3.688 \times 10^{15}$ or $= 4 \times 10^{15}$

AMU-2007

Chemical Equilibrium

229337 The standard emf of a cell, involving one electron change is found to be $0.591 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction is $(F =$ $96500 C {mol}^{- 1}$ )

1

1.0 \times 10^{1}

2

1.0 \times 10^{5}

3

1.0 \times 10^{10}

4

1.0 \times 10^{30}

Explanation:

Equilibrium at $298 K$ is-
$E_{cell}^{o} = \frac{0.0591}{n} \log K_{c}$
$K_{c} =$ Equilibrium constant
Given, $E_{cell}^{o} = 0.591$
Now, $\log K_{c} = \frac{E_{cell}^{o} \times n}{0.0591} = \frac{0.591 \times n}{0.0591}$
$\log K_{c} = 10$
Taking antilog both side, we get-
$K_{c} = 1 \times 10^{10}$

[AIEEE 2004]

Chemical Equilibrium

229333 Which of the following is a wrong statement about equilibrium state?

1 Rate of forward reaction= Rate of backward reaction

2 Equilibrium is dynamic

3 Catalysts increase value of equilibrium constant

4 Free energy change is zero

Explanation:

Chemical equilibrium is dynamic in nature which in other words means that reactions continue to occur in both forward as well as backward direction with the same speed.
Equilibrium constant has definite value for every chemical reaction at a given temperature. It is independent of concentration and catalyst.
Free energy change :
$Δ G = Δ G^{o} + 2.303 RT \log Q$
So, option (a), (b), (d) follows equilibrium state but (c) not follows.
Hence, option (c) is correct.

CG PET -2009

Chemical Equilibrium

229335 The standard emf of a cell, involving one electron change is found to be $0.591 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction is $(F =$ $96500 {Cmol}^{- 1}$ )

1

1.0 \times 10^{1}

2

1.0 \times 10^{5}

3

1.0 \times 10^{10}

4

1.0 \times 10^{30}

Explanation:

Relation between $K_{eq}$ and $E_{cell}$ us
$E_{cell}^{\circ} = \frac{2.303 RT}{nF} \log K_{eq}$
$E_{cell}^{\circ} = \frac{2.303 RT}{nF} \log K_{eq} ($ at $298 K)$
$E_{cell}^{\circ} = \frac{0.0591}{n} \log K_{eq}$
$0.0591 = \frac{0.0591}{n} \log K_{eq}$
$\log K_{eq} = 10 = K_{eq} = antilog 10$
$K_{eq} = 1 \times 10^{10}$

UPTU/UPSEE-2008

Chemical Equilibrium

229336 The equilibrium constant of the reaction
$Cu (s) + 2 {Ag}^{+} (aq) \to {Cu}^{2 +} (aq) + Ag (s)$
[Given : $E^{o} = 0.46 V$ ] will be

1

4 \times 10^{15}

2

4 \times 10^{14}

3

14 \times 10^{15}

4

15.6 \times 10^{5}

Explanation:

The equilibrium constant for the give reaction can be given as, [Given : $E^{o} = 0.46 V$ ]
$\begin{aligned} Cu (s) + 2 {Ag}^{+} (aq) \to {Cu}^{2 +} (aq) + Ag (s) \\ E^{\circ} = \frac{0.0591}{n} \log K \end{aligned}$
Where, $K =$ Equilibrium constant, $n =$ Number of electrons.
On putting the values in eqn, ....(i) we get
$0.46 = \frac{0.0591}{n} \log K$
$\log K = \frac{0.46 \times 2}{0.0591}$
$∴ K =$ Antilog 15.6
$= 3.688 \times 10^{15}$ or $= 4 \times 10^{15}$

AMU-2007

Chemical Equilibrium

229337 The standard emf of a cell, involving one electron change is found to be $0.591 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction is $(F =$ $96500 C {mol}^{- 1}$ )

1

1.0 \times 10^{1}

2

1.0 \times 10^{5}

3

1.0 \times 10^{10}

4

1.0 \times 10^{30}

Explanation:

Equilibrium at $298 K$ is-
$E_{cell}^{o} = \frac{0.0591}{n} \log K_{c}$
$K_{c} =$ Equilibrium constant
Given, $E_{cell}^{o} = 0.591$
Now, $\log K_{c} = \frac{E_{cell}^{o} \times n}{0.0591} = \frac{0.591 \times n}{0.0591}$
$\log K_{c} = 10$
Taking antilog both side, we get-
$K_{c} = 1 \times 10^{10}$

[AIEEE 2004]

Chemical Equilibrium

229333 Which of the following is a wrong statement about equilibrium state?

1 Rate of forward reaction= Rate of backward reaction

2 Equilibrium is dynamic

3 Catalysts increase value of equilibrium constant

4 Free energy change is zero

Explanation:

Chemical equilibrium is dynamic in nature which in other words means that reactions continue to occur in both forward as well as backward direction with the same speed.
Equilibrium constant has definite value for every chemical reaction at a given temperature. It is independent of concentration and catalyst.
Free energy change :
$Δ G = Δ G^{o} + 2.303 RT \log Q$
So, option (a), (b), (d) follows equilibrium state but (c) not follows.
Hence, option (c) is correct.

CG PET -2009

Chemical Equilibrium

229335 The standard emf of a cell, involving one electron change is found to be $0.591 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction is $(F =$ $96500 {Cmol}^{- 1}$ )

1

1.0 \times 10^{1}

2

1.0 \times 10^{5}

3

1.0 \times 10^{10}

4

1.0 \times 10^{30}

Explanation:

Relation between $K_{eq}$ and $E_{cell}$ us
$E_{cell}^{\circ} = \frac{2.303 RT}{nF} \log K_{eq}$
$E_{cell}^{\circ} = \frac{2.303 RT}{nF} \log K_{eq} ($ at $298 K)$
$E_{cell}^{\circ} = \frac{0.0591}{n} \log K_{eq}$
$0.0591 = \frac{0.0591}{n} \log K_{eq}$
$\log K_{eq} = 10 = K_{eq} = antilog 10$
$K_{eq} = 1 \times 10^{10}$

UPTU/UPSEE-2008

Chemical Equilibrium

229336 The equilibrium constant of the reaction
$Cu (s) + 2 {Ag}^{+} (aq) \to {Cu}^{2 +} (aq) + Ag (s)$
[Given : $E^{o} = 0.46 V$ ] will be

1

4 \times 10^{15}

2

4 \times 10^{14}

3

14 \times 10^{15}

4

15.6 \times 10^{5}

Explanation:

The equilibrium constant for the give reaction can be given as, [Given : $E^{o} = 0.46 V$ ]
$\begin{aligned} Cu (s) + 2 {Ag}^{+} (aq) \to {Cu}^{2 +} (aq) + Ag (s) \\ E^{\circ} = \frac{0.0591}{n} \log K \end{aligned}$
Where, $K =$ Equilibrium constant, $n =$ Number of electrons.
On putting the values in eqn, ....(i) we get
$0.46 = \frac{0.0591}{n} \log K$
$\log K = \frac{0.46 \times 2}{0.0591}$
$∴ K =$ Antilog 15.6
$= 3.688 \times 10^{15}$ or $= 4 \times 10^{15}$

AMU-2007

Chemical Equilibrium

229337 The standard emf of a cell, involving one electron change is found to be $0.591 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction is $(F =$ $96500 C {mol}^{- 1}$ )

1

1.0 \times 10^{1}

2

1.0 \times 10^{5}

3

1.0 \times 10^{10}

4

1.0 \times 10^{30}

Explanation:

Equilibrium at $298 K$ is-
$E_{cell}^{o} = \frac{0.0591}{n} \log K_{c}$
$K_{c} =$ Equilibrium constant
Given, $E_{cell}^{o} = 0.591$
Now, $\log K_{c} = \frac{E_{cell}^{o} \times n}{0.0591} = \frac{0.591 \times n}{0.0591}$
$\log K_{c} = 10$
Taking antilog both side, we get-
$K_{c} = 1 \times 10^{10}$

[AIEEE 2004]

Chemical Equilibrium

229333 Which of the following is a wrong statement about equilibrium state?

1 Rate of forward reaction= Rate of backward reaction

2 Equilibrium is dynamic

3 Catalysts increase value of equilibrium constant

4 Free energy change is zero

Explanation:

Chemical equilibrium is dynamic in nature which in other words means that reactions continue to occur in both forward as well as backward direction with the same speed.
Equilibrium constant has definite value for every chemical reaction at a given temperature. It is independent of concentration and catalyst.
Free energy change :
$Δ G = Δ G^{o} + 2.303 RT \log Q$
So, option (a), (b), (d) follows equilibrium state but (c) not follows.
Hence, option (c) is correct.

CG PET -2009

Chemical Equilibrium

229335 The standard emf of a cell, involving one electron change is found to be $0.591 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction is $(F =$ $96500 {Cmol}^{- 1}$ )

1

1.0 \times 10^{1}

2

1.0 \times 10^{5}

3

1.0 \times 10^{10}

4

1.0 \times 10^{30}

Explanation:

Relation between $K_{eq}$ and $E_{cell}$ us
$E_{cell}^{\circ} = \frac{2.303 RT}{nF} \log K_{eq}$
$E_{cell}^{\circ} = \frac{2.303 RT}{nF} \log K_{eq} ($ at $298 K)$
$E_{cell}^{\circ} = \frac{0.0591}{n} \log K_{eq}$
$0.0591 = \frac{0.0591}{n} \log K_{eq}$
$\log K_{eq} = 10 = K_{eq} = antilog 10$
$K_{eq} = 1 \times 10^{10}$

UPTU/UPSEE-2008

Chemical Equilibrium

229336 The equilibrium constant of the reaction
$Cu (s) + 2 {Ag}^{+} (aq) \to {Cu}^{2 +} (aq) + Ag (s)$
[Given : $E^{o} = 0.46 V$ ] will be

1

4 \times 10^{15}

2

4 \times 10^{14}

3

14 \times 10^{15}

4

15.6 \times 10^{5}

Explanation:

The equilibrium constant for the give reaction can be given as, [Given : $E^{o} = 0.46 V$ ]
$\begin{aligned} Cu (s) + 2 {Ag}^{+} (aq) \to {Cu}^{2 +} (aq) + Ag (s) \\ E^{\circ} = \frac{0.0591}{n} \log K \end{aligned}$
Where, $K =$ Equilibrium constant, $n =$ Number of electrons.
On putting the values in eqn, ....(i) we get
$0.46 = \frac{0.0591}{n} \log K$
$\log K = \frac{0.46 \times 2}{0.0591}$
$∴ K =$ Antilog 15.6
$= 3.688 \times 10^{15}$ or $= 4 \times 10^{15}$

AMU-2007

Chemical Equilibrium

229337 The standard emf of a cell, involving one electron change is found to be $0.591 V$ at $25^{\circ} C$ . The equilibrium constant of the reaction is $(F =$ $96500 C {mol}^{- 1}$ )

1

1.0 \times 10^{1}

2

1.0 \times 10^{5}

3

1.0 \times 10^{10}

4

1.0 \times 10^{30}

Explanation:

Equilibrium at $298 K$ is-
$E_{cell}^{o} = \frac{0.0591}{n} \log K_{c}$
$K_{c} =$ Equilibrium constant
Given, $E_{cell}^{o} = 0.591$
Now, $\log K_{c} = \frac{E_{cell}^{o} \times n}{0.0591} = \frac{0.591 \times n}{0.0591}$
$\log K_{c} = 10$
Taking antilog both side, we get-
$K_{c} = 1 \times 10^{10}$

[AIEEE 2004]

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