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11th
Chemistry
Equilibrium
Homogeneous and Heterogeneous Equilibria
$Homogeneous Equilibrium$

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Homogeneous Equilibrium

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Homogeneous equilibrium - definition

In an equilibrium, all the reactants and the products are present in the same phase, it is called homogeneous equilibrium.

H_{2} (g) + I_{2} (g) ⇋ 2 H I (g)

Expression for equilibrium constant - definition

a A (g) + b B (g) ⇋ c C (g) + d D (g)

K_{p} = \frac{p _{A}^{a} p _{B}^{b}}{p _{C}^{c} p _{D}^{d}}

K_{c} = \frac{A ^{a} B ^{b}}{C ^{c} D ^{d}}

Units of equilibrium constant - definition

K_{c} = (m o l L^{- 1})^{Δ n}

K_{c} = (a t m o r b a r)^{Δ n}

where,

Δ n = n_{p r o d u c t} - n_{r e a c t a n t}

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Heterogeneous Equilibrium

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Classification of Chemical Equilibrium

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n mole of $P C l_{3}$ and n mole $C l_{2}$ are allowed to react at constant temperature T to have a total equilibrium pressure $P$ , as : $P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$ . If y mole of $P C l_{3}$ are formed at equilibrium, find $k_{p}$ for the given reaction.

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The reaction, $A + 2 B ⇌ 2 C + D$ was studied using an initial concentration of $B$ which was 1.5 times that of $A$ . But the equilibrium concentration of $A$ and $C$ were found to be equal. Then the $K_{c}$ for the equilibrium is :

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For the reaction: $S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ 2 S O_{3} (g)$
if $K_{p} = K_{c} (R T)^{x}$ ; where the symbols have usual meanings then the value of $x$ is: (assuming ideality)

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The correct relation between $K_{p}$ and $K_{c}$ for the given reaction is:
$H_{2} (g) + l_{1} (g) ⇌ 2 H l (g)$

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For the following gaseous reaction $H_{2} + I_{2} ⇌ 2 H I$ , the equilibrium constant:

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Statement 1: A large value for the equilibrium constant indicates that products are favored at equilibrium.
Statement 2: The ratio of products to reactants at equilibrium is always greater than one.

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$2 N O B r (g) ⇌ 2 N O (g) + B r_{2} (g) .$ If nitrosyl bromide (NOBr) is 33.33 per dissociated at $2 5^{\circ} C$ & a total pressure of 0.28 atm. Calculate $K_{p}$ for the dissociation at this temperature.

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$\frac{K _{c}}{K _{p}}$ for reaction :
$C O (g) + \frac{1}{2} O_{2} (g) ⇋ C O_{2} (g)$ is

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4 moles of $H I$ is taken in a liter closed vessel and heated to equilibrium. At equilibrium, the concentration of $H_{2}$ is one $m o l l i t^{- 1}$ .

Then the equilibrium constant is:

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Equal number of moles of $A$ and $B$ are allowed to react with each till it reaches equilibrium
$2 A + B ⇌ C + D$
The value of $K_{c}$ for this equilibrium can never be

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Homogeneous equilibrium - definition

Expression for equilibrium constant - definition

Units of equilibrium constant - definition

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Related Concepts

Heterogeneous Equilibrium

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Classification of Chemical Equilibrium

Related questions

n mole of PCl3​ and n mole Cl2​ are allowed to react at constant temperature T to have a total equilibrium pressure P, as :PCl3​(g)+Cl2​(g)⇌PCl5​(g). If y mole of PCl3​ are formed at equilibrium, find kp​ for the given reaction.

The reaction, A+2B⇌2C+D was studied using an initial concentration of B which was 1.5 times that of A. But the equilibrium concentration of A and C were found to be equal. Then the Kc​ for the equilibrium is :

For the reaction: SO2​(g)+21​O2​(g)⇌2SO3​(g) if Kp​=Kc​(RT)x; where the symbols have usual meanings then the value of x is: (assuming ideality)

The correct relation between Kp​ and Kc​ for the given reaction is: H2​(g)+l1​(g)⇌2Hl(g)

For the following gaseous reaction H2​+I2​⇌2HI , the equilibrium constant:

Statement 1: A large value for the equilibrium constant indicates that products are favored at equilibrium. Statement 2: The ratio of products to reactants at equilibrium is always greater than one.

2NOBr(g)⇌2NO(g)+Br2​(g). If nitrosyl bromide (NOBr) is 33.33 per dissociated at 25∘C& a total pressure of 0.28 atm. Calculate Kp​ for the dissociation at this temperature.

Kp​Kc​​ for reaction : CO(g)+21​O2​(g)⇋CO2​(g) is

4 moles of HI is taken in a liter closed vessel and heated to equilibrium. At equilibrium, the concentration of H2​ is one mol lit−1. Then the equilibrium constant is:

Equal number of moles of A and B are allowed to react with each till it reaches equilibrium 2A+B⇌C+D The value of Kc​ for this equilibrium can never be

n mole of $P C l_{3}$ and n mole $C l_{2}$ are allowed to react at constant temperature T to have a total equilibrium pressure $P$ , as : $P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$ . If y mole of $P C l_{3}$ are formed at equilibrium, find $k_{p}$ for the given reaction.

The reaction, $A + 2 B ⇌ 2 C + D$ was studied using an initial concentration of $B$ which was 1.5 times that of $A$ . But the equilibrium concentration of $A$ and $C$ were found to be equal. Then the $K_{c}$ for the equilibrium is :

For the reaction: $S O_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ 2 S O_{3} (g)$
if $K_{p} = K_{c} (R T)^{x}$ ; where the symbols have usual meanings then the value of $x$ is: (assuming ideality)

The correct relation between $K_{p}$ and $K_{c}$ for the given reaction is:
$H_{2} (g) + l_{1} (g) ⇌ 2 H l (g)$

For the following gaseous reaction $H_{2} + I_{2} ⇌ 2 H I$ , the equilibrium constant:

Statement 1: A large value for the equilibrium constant indicates that products are favored at equilibrium.
Statement 2: The ratio of products to reactants at equilibrium is always greater than one.

$2 N O B r (g) ⇌ 2 N O (g) + B r_{2} (g) .$ If nitrosyl bromide (NOBr) is 33.33 per dissociated at $2 5^{\circ} C$ & a total pressure of 0.28 atm. Calculate $K_{p}$ for the dissociation at this temperature.

$\frac{K _{c}}{K _{p}}$ for reaction :
$C O (g) + \frac{1}{2} O_{2} (g) ⇋ C O_{2} (g)$ is

4 moles of $H I$ is taken in a liter closed vessel and heated to equilibrium. At equilibrium, the concentration of $H_{2}$ is one $m o l l i t^{- 1}$ .

Then the equilibrium constant is:

Equal number of moles of $A$ and $B$ are allowed to react with each till it reaches equilibrium
$2 A + B ⇌ C + D$
The value of $K_{c}$ for this equilibrium can never be