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Homogeneous and Heterogeneous Equilibria

Equilibrium

Chemistry

Class 11

Overview

2 mins read

Homogeneous Equilibrium

Heterogeneous Equilibrium

If $A g^{+} + 2 N H_{3} ⇌ A g (N H_{3})_{2}^{+}; K_{1} = 1.8 \times 1 0^{7}$
$A g^{+} C l^{-} ⇌ A g C l; K_{2} = 5.6 \times 1 0^{9}$
Then for $A g C l + 2 N H_{3} ⇌ [A g (N H_{3})_{2}]^{+} + C l^{-}$
Equilibrium constant will be

When sulphur (in the form of $S_{8}$ ) is heated at temperature T, at equilibrium, the pressure of $S_{8}$ falls by $30 %$ from 1.0 atm, because $S_{8 (g)}$ is partially converted into $S_{2 (g)}$ . Find the value of $K_{p}$ for this reaction.

$N_{2} O_{4}$ is 25% dissociated at $3 7^{o} C$ and one atmosphere pressure. Calculate $K_{p}$ and the percentage dissociation at $0.1$ atmosphere and $3 7^{o} C$ .

Unit of equilibrium constant is:

The equilibrium constant can be related with energy change as $\frac{d}{d T} ln K_{c} = Δ H / R T^{2}$ .
If true enter 1, if false enter 0.

We know that the relationship between $K_{c}$ and $K_{p}$ is
$K_{p} = K_{c} (R T)^{Δ n}$
What would be the value of $Δ n$ for the reaction,
$N H_{4} C l (s) ⇋ N H_{3} (g) + H C l (g)$

The volume of container in which gas is heated is :

For the reaction :
$2 N O_{2 (g)} ⇌ 2 N O_{(g)} + O_{2 (g)}, (K_{c} = 1.8 \times 1 0^{- 6}$ at $18 4^{o} C) (R = 0.082 L a t m / (m o l . K)$

When $K_{p}$ and $K_{c}$ are compared at $18 4^{o} C$ , it is found that

For the reaction $C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g), K_{p} / K_{c}$ equal to

The equilibrium constant for the reaction $H_{2} (g () + I_{2} (g) ⇌ 2 H I (g)$ is 50 at 500 K. In presence of suitable catalyst, equilibrium is attained four times faster. The equilibrium constant of the reaction at 500 K in the presence of catalyst should be-

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Homogeneous and Heterogeneous Equilibria

Overview

Homogeneous Equilibrium

Heterogeneous Equilibrium

If Ag++2NH3​⇌Ag(NH3​)2+​;K1​=1.8×107 Ag+Cl−⇌AgCl;K2​=5.6×109 Then for AgCl+2NH3​⇌[Ag(NH3​)2​]++Cl− Equilibrium constant will be

When sulphur (in the form of S8​) is heated at temperature T, at equilibrium, the pressure of S8​ falls by 30% from 1.0 atm, because S8(g)​ is partially converted into S2(g)​. Find the value of Kp​ for this reaction.

N2​O4​ is 25% dissociated at 37oC and one atmosphere pressure. Calculate Kp​ and the percentage dissociation at 0.1 atmosphere and 37oC.

Unit of equilibrium constant is:

The equilibrium constant can be related with energy change as dTd​lnKc​=ΔH/RT2. If true enter 1, if false enter 0.

We know that the relationship between Kc​ and Kp​ is Kp​=Kc​(RT)Δn What would be the value of Δn for the reaction, NH4​Cl(s)⇋NH3​(g)+HCl(g)

The volume of container in which gas is heated is :

For the reaction : 2NO2(g)​⇌2NO(g)​+O2(g)​,(Kc​=1.8×10−6 at 184oC)(R=0.082Latm/(mol.K) When Kp​ and Kc​ are compared at 184oC, it is found that

For the reaction CO(g)+Cl2​(g)⇌COCl2​(g),Kp​/Kc​ equal to

The equilibrium constant for the reaction H2​(g()+I2​(g)⇌2HI(g) is 50 at 500 K. In presence of suitable catalyst, equilibrium is attained four times faster. The equilibrium constant of the reaction at 500 K in the presence of catalyst should be-

If $A g^{+} + 2 N H_{3} ⇌ A g (N H_{3})_{2}^{+}; K_{1} = 1.8 \times 1 0^{7}$
$A g^{+} C l^{-} ⇌ A g C l; K_{2} = 5.6 \times 1 0^{9}$
Then for $A g C l + 2 N H_{3} ⇌ [A g (N H_{3})_{2}]^{+} + C l^{-}$
Equilibrium constant will be

When sulphur (in the form of $S_{8}$ ) is heated at temperature T, at equilibrium, the pressure of $S_{8}$ falls by $30 %$ from 1.0 atm, because $S_{8 (g)}$ is partially converted into $S_{2 (g)}$ . Find the value of $K_{p}$ for this reaction.

$N_{2} O_{4}$ is 25% dissociated at $3 7^{o} C$ and one atmosphere pressure. Calculate $K_{p}$ and the percentage dissociation at $0.1$ atmosphere and $3 7^{o} C$ .

The equilibrium constant can be related with energy change as $\frac{d}{d T} ln K_{c} = Δ H / R T^{2}$ .
If true enter 1, if false enter 0.

We know that the relationship between $K_{c}$ and $K_{p}$ is
$K_{p} = K_{c} (R T)^{Δ n}$
What would be the value of $Δ n$ for the reaction,
$N H_{4} C l (s) ⇋ N H_{3} (g) + H C l (g)$

For the reaction :
$2 N O_{2 (g)} ⇌ 2 N O_{(g)} + O_{2 (g)}, (K_{c} = 1.8 \times 1 0^{- 6}$ at $18 4^{o} C) (R = 0.082 L a t m / (m o l . K)$

When $K_{p}$ and $K_{c}$ are compared at $18 4^{o} C$ , it is found that

For the reaction $C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g), K_{p} / K_{c}$ equal to

The equilibrium constant for the reaction $H_{2} (g () + I_{2} (g) ⇌ 2 H I (g)$ is 50 at 500 K. In presence of suitable catalyst, equilibrium is attained four times faster. The equilibrium constant of the reaction at 500 K in the presence of catalyst should be-