Question

Consider the following system at equilibrium where Kc = 1.29×10-2 and H° = 108 kJ/mol at 600 K.

COCl2 (g) CO (g) + Cl2 (g)

The production of CO (g) is favored by:

Indicate True (T) or False (F) for each of the following:

___
T
F
1. decreasing the temperature.
___
T
F
2. increasing the pressure (by changing the volume).
___
T
F
3. decreasing the volume.
___
T
F
4. adding COCl2 .
___
T
F
5. removing Cl2 .


Consider the following system at equilibrium where Kc = 77.5 and H° = -108 kJ/mol at 600 K.

CO (g) + Cl2 (g) COCl2 (g)

The production of COCl2 (g) is favored by:

Indicate True (T) or False (F) for each of the following:

___
T
F
1. decreasing the temperature.
___
T
F
2. decreasing the pressure (by changing the volume).
___
T
F
3. decreasing the volume.
___
T
F
4. removing COCl2 .
___
T
F
5. removing Cl2 .

Answer 1

Part.A

The given reaction is :

COCl₂(g) <----------------> CO(g) + Cl₂(g) , ΔH⁰ = 108 KJ/mol

Now

Expression of Equilibrium constant i.e. K_c = [CO (g) ] [Cl_{2 (g)}] / [COCl_{2 (g)}]

This is an endothermic reaction as ΔH⁰ = +ve

(1). decreasing the temperature (FALSE) , because according to Le-chatelier's principle on decreasing the temperature for endothermic reaction, equilibrium will shifts in backward direction i.e. reactant side and vice-versa. Hence this will not favor production of CO.

(2). increasing the pressure (by changing the volume) (FALSE) because according to Le-chatelier's principle on increases the pressure, equilibrium will shifts in that direction where number of moles of gases are less. Since number of moles of gases are less in reactant side, therefore equilibrium will shifts in reactants side. Hence this will not favor production of CO.

(3). decreasing the volume (FALSE), because decreasing the volume means increasing the pressure. Hence this will not favor production as equilibrium will shifts in reactants side.

(4). adding COCl₂(TRUE) , because according to Le-chatelier's principle on increasing the concentration of reactants or decreasing the concentration of products, equilibrium will shifts in forwards direction i.e. products side and vice-versa. Hence this will favor production of CO.

(5). removing Cl₂(TRUE) , because according to Le-chatelier's principle on increasing the concentration of reactants or decreasing the concentration of products, equilibrium will shifts in forwards direction i.e. products side and vice-versa. Hence this will favor production of CO.

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Part.B

The given reaction is :

CO(g) + Cl₂(g) <----------------> COCl₂(g) , ΔH⁰ = - 108 KJ/mol

Now

Expression of Equilibrium constant i.e. K_c = [COCl_{2 (g)}] /  [CO (g) ] [Cl_{2 (g)}]

This is an exothermic reaction as ΔH⁰ = -ve

(1). decreasing the temperature (TRUE) , because according to Le-chatelier's principle on decreasing the temperature for exothermic reaction, equilibrium will shifts in forward direction i.e. products side and vice-versa. Hence this will favor production of CO.

(2). increasing the pressure (by changing the volume) (TRUE) because according to Le-chatelier's principle on increases the pressure, equilibrium will shifts in that direction where number of moles of gases are less. Since number of moles of gases are less in products side, therefore equilibrium will shifts in products side. Hence this will favor production of CO.

(3). decreasing the volume (TRUE), because decreasing the volume means increasing the pressure. Hence this will favor production as equilibrium will shifts in products side.

(4). adding COCl₂(FALSE) , because according to Le-chatelier's principle on increasing the concentration of products or decreasing the concentration of reactants, equilibrium will shifts in backward direction i.e. reactant side and vice-versa. Hence this will not favor production of CO.

(5). removing Cl₂(FALSE) , because according to Le-chatelier's principle on increasing the concentration of products or decreasing the concentration of reactants, equilibrium will shifts in backward direction i.e. reactant side and vice-versa. Hence this will not favor production of CO.