- Calcium Carbonate (limestone) can decompose according to the following reactio

Question

- Calcium Carbonate (limestone) can decompose according to the following reaction. Supposed the Kp at a certain high temperature is 1.20. What will be the equilibrium pressure of CO2 (g) if a sufficiently sizeable slab of limestone is placed in a reaction chamber?

CaCO3 (s) >< CaO (s) + CO2 (g)

- Carbon monoxide can be decomposed at a reasonable reate in the presence of a transition metal catalyst ( such as iron or nickel ) according to the following exothermic reaction:

2CO (g) >< CO2 (g) + C(s)

Suppose this reaction is at equilbrium. Which way will reaction shift if:

a- The concentration of CO (g) is increased

b- The concentration of CO2 (g) is decreased

c- More solid carbon is added

d- Tempertature is increased

e- Temperature is decreased

f- A piston decreases the volume of the reaction container

g- A piston increases the volume of the reaction container

h- An additional cataylst is added

Explanation / Answer

1)
CaCO3 and CaO are solid.
So, they will not appear in Kp expression
usE:
Kp = p(CO2)
1.20 = p(CO2)
p(CO2) = 1.20 atm

Answer: 1.20 atm

2)
a)
Adding reactant will shift the reaction in the direction of product as per Le chatelier Principle
So, Equilibrium moves to product side

b)
Removing product will shift the reaction in the direction of product as per Le chatelier Principle
So, Equilibrium moves to product side

c)
Adding solid or liquid doesn't affect equilibrium
So, No effect on equilibrium

d)
Increasing Temperature will shift the reaction in a direction which absorbs heat as per Le chatelier Principle
Forward reaction is exothermic in nature
hence, backward reaction will be favoured
So, Equilibrium moves to reactant side

e)
Decreasing Temperature will shift the reaction in a direction which release heat as per Le chatelier Principle
Forward reaction is exothermic in nature
hence, forward reaction will be favoured
So, Equilibrium moves to product side

f)
Decreasing volume will increase the pressure which in turn shift the reaction in a direction which have lesser gaseous molecules as per Le chatelier Principle
Here product has less gaseous molecule
So equilibrium will move to right
So, Equilibrium moves to product side

g)
Increasing volume will decrease the pressure which in turn shift the reaction in a direction which have greater gaseous molecules as per Le chatelier Principle
Here reactant has more gaseous molecule
So equilibrium will move to left
So, Equilibrium moves to reactant side

h)
Catalyst doesn't affect equilibrium
So, No effect on equilibrium

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.