Equilibrium Test - 53

$$\because Salts\ NH_{4}A, NH_{4}B, NH_{4}C$$ are of weak acid and weak base.

The equilibrium, $$S{ O }_{ 2 }{ Cl }_{ 2 }(g)\rightleftharpoons S{ O }_{ 2 }(g)+{ Cl }_{ 2 }(g)$$ is attained at $${25}^{o}C$$ in a closed container and an inert gas, helium, is introduced.  

With increase in temperature, the value of the equilibrium constant decreases.
As temperature increases, the amount of products decreases and the amount of reactants increases.
With increase in temperature, the equilibrium shifts to left.
This is characteristic of exothermic reaction.

For the reaction,
$$P{ Cl }_{ 5 }(g)\rightleftharpoons P{ Cl }_{ 3 }(g)+{ Cl }_{ 2 }(g)$$
The forward reaction at constant temperature is favoured by:
Introducing inert gas at constant pressure, introducing $$ \displaystyle P{ Cl }_{ 5 }$$ gas, removing chlorine gas and removing $$ \displaystyle P{ Cl }_{ 3 }$$ gas.
Note: Introducing inert gas at constant volume has no effect on equilibrium.

 With increase in temperature, the value of the equilibrium constant decreases.
As temperature increases, the amount of products decreases and the amount of reactants increases.
With increase in temperature, the equilibrium shifts to left.
This is characteristic of exothermic reaction.

Given, Activation energy of forward reaction, $${ E }_{ f }=12KJ/mol$$

In a flask colourless $${N}_{2}{O}_{4}$$ is in equilibrium with brown coloured $${NO}_{2}$$. At equilibrium, when the flask is heated to $${100}^{o}C$$ the brown colour deepens and on cooling, the brown colour became less coloured. The change in enthalpy $$\Delta H$$ of the system is positive.
Colourless $$  \displaystyle {N}_{2}{O}_{4} \rightleftharpoons $$ brown $$ \displaystyle {NO}_{2}$$
When temperature is increased (by heating the flask), brown colour deepens, so more and more $$ \displaystyle {NO}_{2}$$ is formed and forward reaction is favoured.
When temperature is decreased (by cooling the flask), brown colour became less coloured, so more and more $$ \displaystyle {NO}_{2}$$ is converted to $${N}_{2}{O}_{4}$$ and reverse reaction is favoured.
This is characteristic of endothermic reaction with positive value of enthalpy change.

To the system
$$La{ Cl }_{ 3 }(s)+{ H }_{ 2 }O(g)\rightleftharpoons LaClO(s)+2HCl(g)$$- heat already at equilibrium, more water vapour is added without altering $$T$$ or $$V$$ of the system. When equilibrium re-established, the pressure of water vapour is doubled. The pressure of $$HCl$$ present in the system increases by a factor of $${2}^{1/2}$$.

The equilibrium constant expression is
$$ \displaystyle K= \dfrac {[HCl]^2}{[H2O]}$$

Before addition of water vapour,
$$ \displaystyle K= \dfrac {[HCl]_i^2}{[H2O]_i}$$......(1)

After addition of water vapour,
$$ \displaystyle K= \dfrac {[HCl]_f^2}{[H2O]_f}$$......(2)

But (1) $$ \displaystyle =$$ (2)
$$ \displaystyle \dfrac {[HCl]_i^2}{[H2O]_i}=\dfrac {[HCl]_f^2}{[H2O]_f}$$......(3)

 When equilibrium re-established, the pressure of water vapour is doubled.
$$ \displaystyle [H2O]_f=2[H2O]_i$$......(4)

Substitute (4) in (3)
$$ \displaystyle \dfrac {[HCl]_i^2}{[H2O]_i}=\dfrac {[HCl]_f^2} { (2[H2O]_i)}$$
$$ \displaystyle  {[HCl]_i^2}= {[HCl]_f^2} \times \dfrac {1}{2}$$
$$ \displaystyle  {[HCl]_i}= {[HCl]_f} \times \sqrt {\dfrac {1}{2}}$$

$$ \displaystyle [HCl]_f={\sqrt {2}}\times [HCl]_i$$

$${Cu}^{2+}$$ ions react with $${Fe}^{2+}$$ ions according to the following reaction:
$${ Cu }^{ 2+ }+2{ Fe }^{ 2+ }\rightleftharpoons Cu+2{ Fe }^{ 3+ }$$
At equilibrium, the concentration of $${Cu}^{2+}$$ ions is not changed by the addition of $$Cu$$.
$$Cu$$ is a solid and it's concentration does not appear in the equilibrium constant expression.