QBManager

Chemical Equilibrium

229274 For a chemical reaction $A + B C$ , the thermodynamic equilibrium constant $K_{p}$ is

1 in

{atm}^{- 2}

2 in

{atm}^{- 3}

3 in

{atm}^{- 1}

4 dimensionless

Explanation:

For the reaction,
$A + B ⇌ C$
$Δ n_{g} = 1 - 2 = - 1$
$K_{p} = (atmosphere)^{- 1}$

JCECE - 2011

Chemical Equilibrium

229281 For the decomposition reaction of lime stone, the correct expression for rate constant, $K_{p}$ is

1

K_{p} = \frac{P_{CaO} \cdot P_{{CO}_{2}}}{P_{{CaCO}_{3}}}

2

K_{p} = \frac{[CaO] [{CO}_{2}]}{[{CaCO}_{3}]}

3

K_{p} = \frac{P_{CaO} + P_{{CO}_{2}}}{P_{{CaCO}_{3}}}

4

K_{p} = P_{{CO}_{2}}

Explanation:

The decomposition reaction of ${CaCO}_{3}$ is ${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g)$
Thus, $K_{p} = P_{{CO}_{2}}$
As concentration of solid remain unchanged it is taken as unity.

UPTU/UPSEE-2010

Chemical Equilibrium

229292 At constant temperature, the equilibrium constant $(K_{p})$ for decomposition reaction $N_{2} O_{4}$ $2 {NO}_{2}$ is expressed by $K_{p} = (4 x^{2} P) / (1 - x^{2})$ , where, $P =$ pressure; $x =$ extent of decomposition, which one of the following statement is true?

1

K_{p}

increases with increase of

P

?

2

K_{p}

increases with increase of

x

.

3

K_{p}

increases with decrease of

x

.

4

K_{p}

remains constant with change in

P

and

x

.

Explanation:

At constant temperature, the equilibrium constant $(K_{p})$ for the decomposition reaction $N_{2} O_{4} ⇌$
$2 {NO}_{2}$ is expressed by $K_{p} = \frac{(4 x^{2} P)}{(1 - x^{2})}$ , where $P =$ pressure and $x =$ extent of decomposition. $K_{p}$ remains constant with change in $P$ and $x$ . The value of the equilibrium constant is independent of the partial pressure or degree of dissociation.

SRMJEEE - 2008

Chemical Equilibrium

229285 In an equilibrium reaction, if $Δ n_{g}$ is positive then

1

K_{c} = K_{p}

2

K_{c} < K_{p}

3

K_{c} > K_{p}

4

K_{c} = 0

Explanation:

As we know the relation between $K_{p}$ and $K_{c}$ is -
$\begin{matrix} K_{p} = K_{c} (RT)^{Δ n_{g}} \\ If, Δ n_{g} = + ve (Given) \end{matrix}$
Then, $K_{p} > K_{c}$

SRMJEEE - 2009

Chemical Equilibrium

229274 For a chemical reaction $A + B C$ , the thermodynamic equilibrium constant $K_{p}$ is

1 in

{atm}^{- 2}

2 in

{atm}^{- 3}

3 in

{atm}^{- 1}

4 dimensionless

Explanation:

For the reaction,
$A + B ⇌ C$
$Δ n_{g} = 1 - 2 = - 1$
$K_{p} = (atmosphere)^{- 1}$

JCECE - 2011

Chemical Equilibrium

229281 For the decomposition reaction of lime stone, the correct expression for rate constant, $K_{p}$ is

1

K_{p} = \frac{P_{CaO} \cdot P_{{CO}_{2}}}{P_{{CaCO}_{3}}}

2

K_{p} = \frac{[CaO] [{CO}_{2}]}{[{CaCO}_{3}]}

3

K_{p} = \frac{P_{CaO} + P_{{CO}_{2}}}{P_{{CaCO}_{3}}}

4

K_{p} = P_{{CO}_{2}}

Explanation:

The decomposition reaction of ${CaCO}_{3}$ is ${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g)$
Thus, $K_{p} = P_{{CO}_{2}}$
As concentration of solid remain unchanged it is taken as unity.

UPTU/UPSEE-2010

Chemical Equilibrium

229292 At constant temperature, the equilibrium constant $(K_{p})$ for decomposition reaction $N_{2} O_{4}$ $2 {NO}_{2}$ is expressed by $K_{p} = (4 x^{2} P) / (1 - x^{2})$ , where, $P =$ pressure; $x =$ extent of decomposition, which one of the following statement is true?

1

K_{p}

increases with increase of

P

?

2

K_{p}

increases with increase of

x

.

3

K_{p}

increases with decrease of

x

.

4

K_{p}

remains constant with change in

P

and

x

.

Explanation:

At constant temperature, the equilibrium constant $(K_{p})$ for the decomposition reaction $N_{2} O_{4} ⇌$
$2 {NO}_{2}$ is expressed by $K_{p} = \frac{(4 x^{2} P)}{(1 - x^{2})}$ , where $P =$ pressure and $x =$ extent of decomposition. $K_{p}$ remains constant with change in $P$ and $x$ . The value of the equilibrium constant is independent of the partial pressure or degree of dissociation.

SRMJEEE - 2008

Chemical Equilibrium

229285 In an equilibrium reaction, if $Δ n_{g}$ is positive then

1

K_{c} = K_{p}

2

K_{c} < K_{p}

3

K_{c} > K_{p}

4

K_{c} = 0

Explanation:

As we know the relation between $K_{p}$ and $K_{c}$ is -
$\begin{matrix} K_{p} = K_{c} (RT)^{Δ n_{g}} \\ If, Δ n_{g} = + ve (Given) \end{matrix}$
Then, $K_{p} > K_{c}$

SRMJEEE - 2009

Chemical Equilibrium

229274 For a chemical reaction $A + B C$ , the thermodynamic equilibrium constant $K_{p}$ is

1 in

{atm}^{- 2}

2 in

{atm}^{- 3}

3 in

{atm}^{- 1}

4 dimensionless

Explanation:

For the reaction,
$A + B ⇌ C$
$Δ n_{g} = 1 - 2 = - 1$
$K_{p} = (atmosphere)^{- 1}$

JCECE - 2011

Chemical Equilibrium

229281 For the decomposition reaction of lime stone, the correct expression for rate constant, $K_{p}$ is

1

K_{p} = \frac{P_{CaO} \cdot P_{{CO}_{2}}}{P_{{CaCO}_{3}}}

2

K_{p} = \frac{[CaO] [{CO}_{2}]}{[{CaCO}_{3}]}

3

K_{p} = \frac{P_{CaO} + P_{{CO}_{2}}}{P_{{CaCO}_{3}}}

4

K_{p} = P_{{CO}_{2}}

Explanation:

The decomposition reaction of ${CaCO}_{3}$ is ${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g)$
Thus, $K_{p} = P_{{CO}_{2}}$
As concentration of solid remain unchanged it is taken as unity.

UPTU/UPSEE-2010

Chemical Equilibrium

229292 At constant temperature, the equilibrium constant $(K_{p})$ for decomposition reaction $N_{2} O_{4}$ $2 {NO}_{2}$ is expressed by $K_{p} = (4 x^{2} P) / (1 - x^{2})$ , where, $P =$ pressure; $x =$ extent of decomposition, which one of the following statement is true?

1

K_{p}

increases with increase of

P

?

2

K_{p}

increases with increase of

x

.

3

K_{p}

increases with decrease of

x

.

4

K_{p}

remains constant with change in

P

and

x

.

Explanation:

At constant temperature, the equilibrium constant $(K_{p})$ for the decomposition reaction $N_{2} O_{4} ⇌$
$2 {NO}_{2}$ is expressed by $K_{p} = \frac{(4 x^{2} P)}{(1 - x^{2})}$ , where $P =$ pressure and $x =$ extent of decomposition. $K_{p}$ remains constant with change in $P$ and $x$ . The value of the equilibrium constant is independent of the partial pressure or degree of dissociation.

SRMJEEE - 2008

Chemical Equilibrium

229285 In an equilibrium reaction, if $Δ n_{g}$ is positive then

1

K_{c} = K_{p}

2

K_{c} < K_{p}

3

K_{c} > K_{p}

4

K_{c} = 0

Explanation:

As we know the relation between $K_{p}$ and $K_{c}$ is -
$\begin{matrix} K_{p} = K_{c} (RT)^{Δ n_{g}} \\ If, Δ n_{g} = + ve (Given) \end{matrix}$
Then, $K_{p} > K_{c}$

SRMJEEE - 2009

Chemical Equilibrium

229274 For a chemical reaction $A + B C$ , the thermodynamic equilibrium constant $K_{p}$ is

1 in

{atm}^{- 2}

2 in

{atm}^{- 3}

3 in

{atm}^{- 1}

4 dimensionless

Explanation:

For the reaction,
$A + B ⇌ C$
$Δ n_{g} = 1 - 2 = - 1$
$K_{p} = (atmosphere)^{- 1}$

JCECE - 2011

Chemical Equilibrium

229281 For the decomposition reaction of lime stone, the correct expression for rate constant, $K_{p}$ is

1

K_{p} = \frac{P_{CaO} \cdot P_{{CO}_{2}}}{P_{{CaCO}_{3}}}

2

K_{p} = \frac{[CaO] [{CO}_{2}]}{[{CaCO}_{3}]}

3

K_{p} = \frac{P_{CaO} + P_{{CO}_{2}}}{P_{{CaCO}_{3}}}

4

K_{p} = P_{{CO}_{2}}

Explanation:

The decomposition reaction of ${CaCO}_{3}$ is ${CaCO}_{3} (s) ⇌ CaO (s) + {CO}_{2} (g)$
Thus, $K_{p} = P_{{CO}_{2}}$
As concentration of solid remain unchanged it is taken as unity.

UPTU/UPSEE-2010

Chemical Equilibrium

229292 At constant temperature, the equilibrium constant $(K_{p})$ for decomposition reaction $N_{2} O_{4}$ $2 {NO}_{2}$ is expressed by $K_{p} = (4 x^{2} P) / (1 - x^{2})$ , where, $P =$ pressure; $x =$ extent of decomposition, which one of the following statement is true?

1

K_{p}

increases with increase of

P

?

2

K_{p}

increases with increase of

x

.

3

K_{p}

increases with decrease of

x

.

4

K_{p}

remains constant with change in

P

and

x

.

Explanation:

At constant temperature, the equilibrium constant $(K_{p})$ for the decomposition reaction $N_{2} O_{4} ⇌$
$2 {NO}_{2}$ is expressed by $K_{p} = \frac{(4 x^{2} P)}{(1 - x^{2})}$ , where $P =$ pressure and $x =$ extent of decomposition. $K_{p}$ remains constant with change in $P$ and $x$ . The value of the equilibrium constant is independent of the partial pressure or degree of dissociation.

SRMJEEE - 2008

Chemical Equilibrium

229285 In an equilibrium reaction, if $Δ n_{g}$ is positive then

1

K_{c} = K_{p}

2

K_{c} < K_{p}

3

K_{c} > K_{p}

4

K_{c} = 0

Explanation:

As we know the relation between $K_{p}$ and $K_{c}$ is -
$\begin{matrix} K_{p} = K_{c} (RT)^{Δ n_{g}} \\ If, Δ n_{g} = + ve (Given) \end{matrix}$
Then, $K_{p} > K_{c}$

SRMJEEE - 2009

Question Bank Series

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Question Bank Series

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