For the reaction, "CaCO"_3"(s)" ⇌ "CaO(s) + CO"_2"(g)",what will happen in each of the followig cases?

(a) you add some carbon dioxide (b) you lower the pressure (c) you remove some calcium oxide (d) you increase the volume

Feb 20, 2017

Here's what I get.

Explanation:

This is an exercise on Le Châtelier's Principle, which states that

When a stress is applied to a system at equilibrium, the system will respond in such a way as to relieve the stress.

The system at equilibrium is

$\text{CaCO"_3"(s)" ⇋ "CaO(s)" + "CO"_2"(g)}$

Adding ${\text{CO}}_{2}$

The system will respond to remove ${\text{CO}}_{2}$.

The position of equilibrium will move to the left and the mass of ${\text{CaCO}}_{3}$ will increase.

Lowering the pressure

The system will try to increase the pressure by producing more ${\text{CO}}_{2}$.

The position of equilibrium will move to the right and the mass of ${\text{CaCO}}_{3}$ will decrease.

Removing some $\text{CaO}$

Changing the amount of a solid has no effect on the position of equilibrium.

The mass of ${\text{CaCO}}_{3}$ will be unchanged.

Increasing the volume

Increasing the volume will decrease the pressure of ${\text{CO}}_{2}$.

The system will try to increase the pressure by producing more ${\text{CO}}_{2}$.

The position of equilibrium will move to the right and the mass of ${\text{CaCO}}_{3}$ will decrease.