QBManager

Chemical Equilibrium

229251 The $K_{sp}$ of ${Ag}_{2} {CrO}_{4}, AgCl, AgBr$ and $AgI$ are respectively $1.1 \times 10^{- 12}, 1.8 \times 10^{- 10}, 5.0 \times 10^{- 13}$ , $8.3 \times 10^{- 17}$ . which one of the following salts will precipitate last if ${AgNO}_{3}$ solution is added to the solution containing equal moles of $NaCl$ , $NaBr$ , BaI and ${Na}_{2} {CrO}_{4}$ ?

1

AgBr

2

{Ag}_{2} {CrO}_{4}

3 AgI

4

AgCl

Explanation:

For ${Ag}_{2} {CrO}_{4}$
$\begin{array}{lccc} {Ag}_{2} {CrO}_{4} & \to & 2 {Ag}^{+} & + & {CrO}_{4}^{2 -} \\ Initially & 1 & 0 & 0 \\ At time t & 1 - S & S & S \end{array}$
$K_{sp} = \frac{(2 S)^{2} S}{(1 - S)} as S <<< 1$
So, $K_{sp} = (2 S)^{2} S$
$\begin{aligned} 1.1 \times 10^{- 12} = 4 S^{3} \\ S = 6.5 \times 10^{- 5} \end{aligned}$
For $AgCl$ :-
$\begin{aligned} AgCl \to {Ag}^{+} + {Cl}^{-} \\ S S \\ K_{sp} = S^{2} \\ 1.8 \times 10^{- 10} = S^{2} \\ S = 1.34 \times 10^{- 5} \end{aligned}$
Similarly, for $AgBr$ and $AgI$ is $7.1 \times 10^{- 7}$ and $9.1 \times 10^{- 9}$ respectively. Since, solubility of ${Ag}_{2} {CrO}_{4}$ is the highest, it will precipitate last

NEET-2015

Chemical Equilibrium

229253 For a reaction $C (s) + {CO}_{2} (g) ⇌ 2 CO (g)$ the partial pressures of ${CO}_{2}$ and $CO$ are 4 and 8 atm,
respectively, $K_{p}$ for the reaction is

1 0.5

2 2

3 16

4 4

Explanation:

Given data: $p_{{CO}_{2}} = 4 atm, p_{CO} = 8 atm$
For the reaction-
$∴ \begin{aligned} C (s) & {CO}_{2} (g) ⇌ 2 CO (g) \\ K_{p} & = \frac{{(p_{CO})}^{2}}{p_{{CO}_{2}}} \\ K_{p} & = \frac{(8)^{2}}{4} \\ K_{p} & = \frac{64}{4} \\ K_{p} & = 16 \end{aligned}$

J and K CET-(2015)

Chemical Equilibrium

229254 $K_{p} / K_{c}$ for the reaction,
$A (g) + 2 B (g) ⇌ {AB}_{2} (g) is$

1 RT

2

(RT)^{2}

3

1 / RT

4

1 / (RT)^{2}

Explanation:

For the reaction:
$A + 2 B ⇌ {AB}_{2}$
$∴ K_{p} = K_{c} (RT)^{Δ n_{g}}$
$Δ n_{g} = 1 - (1 + 2) = - 2$
So,
$K_{p} = K_{c} (RT)^{- 2}$
$\frac{K_{p}}{K_{c}} = \frac{1}{(RT)^{2}}$

J and K CET-(2015)

Chemical Equilibrium

229256 Which of the following statements is correct for a reversible process in a state of equilibrium ?

1

Δ G^{o} = - 2.30 RT \log K

2

Δ G^{o} = 2.30 RT \log K

3

Δ G = - 2.30 RT \log K

4

Δ G = 2.30 RT \log K

Explanation:

: A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation-
$Δ G = Δ G^{o} + RT \ln Q \dots . . (i)]$
At equilibrium, when $Δ G = 0$ and $Q = K_{c}$
equation (i) becomes,
$Δ G = Δ G^{o} + RT \ln K_{c} = 0$
$Δ G^{o} = - RT \ln K$
$Δ G^{o} = - 2.303 RT \log K$

NEET-2015

Chemical Equilibrium

229251 The $K_{sp}$ of ${Ag}_{2} {CrO}_{4}, AgCl, AgBr$ and $AgI$ are respectively $1.1 \times 10^{- 12}, 1.8 \times 10^{- 10}, 5.0 \times 10^{- 13}$ , $8.3 \times 10^{- 17}$ . which one of the following salts will precipitate last if ${AgNO}_{3}$ solution is added to the solution containing equal moles of $NaCl$ , $NaBr$ , BaI and ${Na}_{2} {CrO}_{4}$ ?

1

AgBr

2

{Ag}_{2} {CrO}_{4}

3 AgI

4

AgCl

Explanation:

For ${Ag}_{2} {CrO}_{4}$
$\begin{array}{lccc} {Ag}_{2} {CrO}_{4} & \to & 2 {Ag}^{+} & + & {CrO}_{4}^{2 -} \\ Initially & 1 & 0 & 0 \\ At time t & 1 - S & S & S \end{array}$
$K_{sp} = \frac{(2 S)^{2} S}{(1 - S)} as S <<< 1$
So, $K_{sp} = (2 S)^{2} S$
$\begin{aligned} 1.1 \times 10^{- 12} = 4 S^{3} \\ S = 6.5 \times 10^{- 5} \end{aligned}$
For $AgCl$ :-
$\begin{aligned} AgCl \to {Ag}^{+} + {Cl}^{-} \\ S S \\ K_{sp} = S^{2} \\ 1.8 \times 10^{- 10} = S^{2} \\ S = 1.34 \times 10^{- 5} \end{aligned}$
Similarly, for $AgBr$ and $AgI$ is $7.1 \times 10^{- 7}$ and $9.1 \times 10^{- 9}$ respectively. Since, solubility of ${Ag}_{2} {CrO}_{4}$ is the highest, it will precipitate last

NEET-2015

Chemical Equilibrium

229253 For a reaction $C (s) + {CO}_{2} (g) ⇌ 2 CO (g)$ the partial pressures of ${CO}_{2}$ and $CO$ are 4 and 8 atm,
respectively, $K_{p}$ for the reaction is

1 0.5

2 2

3 16

4 4

Explanation:

Given data: $p_{{CO}_{2}} = 4 atm, p_{CO} = 8 atm$
For the reaction-
$∴ \begin{aligned} C (s) & {CO}_{2} (g) ⇌ 2 CO (g) \\ K_{p} & = \frac{{(p_{CO})}^{2}}{p_{{CO}_{2}}} \\ K_{p} & = \frac{(8)^{2}}{4} \\ K_{p} & = \frac{64}{4} \\ K_{p} & = 16 \end{aligned}$

J and K CET-(2015)

Chemical Equilibrium

229254 $K_{p} / K_{c}$ for the reaction,
$A (g) + 2 B (g) ⇌ {AB}_{2} (g) is$

1 RT

2

(RT)^{2}

3

1 / RT

4

1 / (RT)^{2}

Explanation:

For the reaction:
$A + 2 B ⇌ {AB}_{2}$
$∴ K_{p} = K_{c} (RT)^{Δ n_{g}}$
$Δ n_{g} = 1 - (1 + 2) = - 2$
So,
$K_{p} = K_{c} (RT)^{- 2}$
$\frac{K_{p}}{K_{c}} = \frac{1}{(RT)^{2}}$

J and K CET-(2015)

Chemical Equilibrium

229256 Which of the following statements is correct for a reversible process in a state of equilibrium ?

1

Δ G^{o} = - 2.30 RT \log K

2

Δ G^{o} = 2.30 RT \log K

3

Δ G = - 2.30 RT \log K

4

Δ G = 2.30 RT \log K

Explanation:

: A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation-
$Δ G = Δ G^{o} + RT \ln Q \dots . . (i)]$
At equilibrium, when $Δ G = 0$ and $Q = K_{c}$
equation (i) becomes,
$Δ G = Δ G^{o} + RT \ln K_{c} = 0$
$Δ G^{o} = - RT \ln K$
$Δ G^{o} = - 2.303 RT \log K$

NEET-2015

Chemical Equilibrium

229251 The $K_{sp}$ of ${Ag}_{2} {CrO}_{4}, AgCl, AgBr$ and $AgI$ are respectively $1.1 \times 10^{- 12}, 1.8 \times 10^{- 10}, 5.0 \times 10^{- 13}$ , $8.3 \times 10^{- 17}$ . which one of the following salts will precipitate last if ${AgNO}_{3}$ solution is added to the solution containing equal moles of $NaCl$ , $NaBr$ , BaI and ${Na}_{2} {CrO}_{4}$ ?

1

AgBr

2

{Ag}_{2} {CrO}_{4}

3 AgI

4

AgCl

Explanation:

For ${Ag}_{2} {CrO}_{4}$
$\begin{array}{lccc} {Ag}_{2} {CrO}_{4} & \to & 2 {Ag}^{+} & + & {CrO}_{4}^{2 -} \\ Initially & 1 & 0 & 0 \\ At time t & 1 - S & S & S \end{array}$
$K_{sp} = \frac{(2 S)^{2} S}{(1 - S)} as S <<< 1$
So, $K_{sp} = (2 S)^{2} S$
$\begin{aligned} 1.1 \times 10^{- 12} = 4 S^{3} \\ S = 6.5 \times 10^{- 5} \end{aligned}$
For $AgCl$ :-
$\begin{aligned} AgCl \to {Ag}^{+} + {Cl}^{-} \\ S S \\ K_{sp} = S^{2} \\ 1.8 \times 10^{- 10} = S^{2} \\ S = 1.34 \times 10^{- 5} \end{aligned}$
Similarly, for $AgBr$ and $AgI$ is $7.1 \times 10^{- 7}$ and $9.1 \times 10^{- 9}$ respectively. Since, solubility of ${Ag}_{2} {CrO}_{4}$ is the highest, it will precipitate last

NEET-2015

Chemical Equilibrium

229253 For a reaction $C (s) + {CO}_{2} (g) ⇌ 2 CO (g)$ the partial pressures of ${CO}_{2}$ and $CO$ are 4 and 8 atm,
respectively, $K_{p}$ for the reaction is

1 0.5

2 2

3 16

4 4

Explanation:

Given data: $p_{{CO}_{2}} = 4 atm, p_{CO} = 8 atm$
For the reaction-
$∴ \begin{aligned} C (s) & {CO}_{2} (g) ⇌ 2 CO (g) \\ K_{p} & = \frac{{(p_{CO})}^{2}}{p_{{CO}_{2}}} \\ K_{p} & = \frac{(8)^{2}}{4} \\ K_{p} & = \frac{64}{4} \\ K_{p} & = 16 \end{aligned}$

J and K CET-(2015)

Chemical Equilibrium

229254 $K_{p} / K_{c}$ for the reaction,
$A (g) + 2 B (g) ⇌ {AB}_{2} (g) is$

1 RT

2

(RT)^{2}

3

1 / RT

4

1 / (RT)^{2}

Explanation:

For the reaction:
$A + 2 B ⇌ {AB}_{2}$
$∴ K_{p} = K_{c} (RT)^{Δ n_{g}}$
$Δ n_{g} = 1 - (1 + 2) = - 2$
So,
$K_{p} = K_{c} (RT)^{- 2}$
$\frac{K_{p}}{K_{c}} = \frac{1}{(RT)^{2}}$

J and K CET-(2015)

Chemical Equilibrium

229256 Which of the following statements is correct for a reversible process in a state of equilibrium ?

1

Δ G^{o} = - 2.30 RT \log K

2

Δ G^{o} = 2.30 RT \log K

3

Δ G = - 2.30 RT \log K

4

Δ G = 2.30 RT \log K

Explanation:

: A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation-
$Δ G = Δ G^{o} + RT \ln Q \dots . . (i)]$
At equilibrium, when $Δ G = 0$ and $Q = K_{c}$
equation (i) becomes,
$Δ G = Δ G^{o} + RT \ln K_{c} = 0$
$Δ G^{o} = - RT \ln K$
$Δ G^{o} = - 2.303 RT \log K$

NEET-2015

Chemical Equilibrium

229251 The $K_{sp}$ of ${Ag}_{2} {CrO}_{4}, AgCl, AgBr$ and $AgI$ are respectively $1.1 \times 10^{- 12}, 1.8 \times 10^{- 10}, 5.0 \times 10^{- 13}$ , $8.3 \times 10^{- 17}$ . which one of the following salts will precipitate last if ${AgNO}_{3}$ solution is added to the solution containing equal moles of $NaCl$ , $NaBr$ , BaI and ${Na}_{2} {CrO}_{4}$ ?

1

AgBr

2

{Ag}_{2} {CrO}_{4}

3 AgI

4

AgCl

Explanation:

For ${Ag}_{2} {CrO}_{4}$
$\begin{array}{lccc} {Ag}_{2} {CrO}_{4} & \to & 2 {Ag}^{+} & + & {CrO}_{4}^{2 -} \\ Initially & 1 & 0 & 0 \\ At time t & 1 - S & S & S \end{array}$
$K_{sp} = \frac{(2 S)^{2} S}{(1 - S)} as S <<< 1$
So, $K_{sp} = (2 S)^{2} S$
$\begin{aligned} 1.1 \times 10^{- 12} = 4 S^{3} \\ S = 6.5 \times 10^{- 5} \end{aligned}$
For $AgCl$ :-
$\begin{aligned} AgCl \to {Ag}^{+} + {Cl}^{-} \\ S S \\ K_{sp} = S^{2} \\ 1.8 \times 10^{- 10} = S^{2} \\ S = 1.34 \times 10^{- 5} \end{aligned}$
Similarly, for $AgBr$ and $AgI$ is $7.1 \times 10^{- 7}$ and $9.1 \times 10^{- 9}$ respectively. Since, solubility of ${Ag}_{2} {CrO}_{4}$ is the highest, it will precipitate last

NEET-2015

Chemical Equilibrium

229253 For a reaction $C (s) + {CO}_{2} (g) ⇌ 2 CO (g)$ the partial pressures of ${CO}_{2}$ and $CO$ are 4 and 8 atm,
respectively, $K_{p}$ for the reaction is

1 0.5

2 2

3 16

4 4

Explanation:

Given data: $p_{{CO}_{2}} = 4 atm, p_{CO} = 8 atm$
For the reaction-
$∴ \begin{aligned} C (s) & {CO}_{2} (g) ⇌ 2 CO (g) \\ K_{p} & = \frac{{(p_{CO})}^{2}}{p_{{CO}_{2}}} \\ K_{p} & = \frac{(8)^{2}}{4} \\ K_{p} & = \frac{64}{4} \\ K_{p} & = 16 \end{aligned}$

J and K CET-(2015)

Chemical Equilibrium

229254 $K_{p} / K_{c}$ for the reaction,
$A (g) + 2 B (g) ⇌ {AB}_{2} (g) is$

1 RT

2

(RT)^{2}

3

1 / RT

4

1 / (RT)^{2}

Explanation:

For the reaction:
$A + 2 B ⇌ {AB}_{2}$
$∴ K_{p} = K_{c} (RT)^{Δ n_{g}}$
$Δ n_{g} = 1 - (1 + 2) = - 2$
So,
$K_{p} = K_{c} (RT)^{- 2}$
$\frac{K_{p}}{K_{c}} = \frac{1}{(RT)^{2}}$

J and K CET-(2015)

Chemical Equilibrium

229256 Which of the following statements is correct for a reversible process in a state of equilibrium ?

1

Δ G^{o} = - 2.30 RT \log K

2

Δ G^{o} = 2.30 RT \log K

3

Δ G = - 2.30 RT \log K

4

Δ G = 2.30 RT \log K

Explanation:

: A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation-
$Δ G = Δ G^{o} + RT \ln Q \dots . . (i)]$
At equilibrium, when $Δ G = 0$ and $Q = K_{c}$
equation (i) becomes,
$Δ G = Δ G^{o} + RT \ln K_{c} = 0$
$Δ G^{o} = - RT \ln K$
$Δ G^{o} = - 2.303 RT \log K$

NEET-2015

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