# Imagine that the system is allowed to come to equilibrium. How would each of the following changes affect the equilibrium? Indicate whether the change would favor the formation of more products, more reactants, or would not lead to any change?

## Sep 22, 2016

Here is my reasoning.

Let's treat the heat as a product and rewrite the equation as

$\text{Ca"("NO"_3)_2"(aq)" + "CO"_2"(g)" + "2NaOH(aq)" ⇌ "CaCO"_3"(s)" + "2NaNO"_3"(aq)" + "H"_2"O(l) + heat}$

Now, we can apply Le Châtelier's Principle to answer the questions.

A. Increasing the temperature

Adding more heat (on the right) causes the position of equilibrium to shift to the left.

The change favours the formation of more reactant.

B. Increasing the partial pressure of ${\text{CO}}_{2}$

Increasing the partial pressure of ${\text{CO}}_{2}$ (on the left) causes the position of equilibrium to shift to the right.

The change favours the formation of more product.

C. Adding more $\text{NaOH}$

Adding more $\text{NaOH}$ (on the left) causes the position of equilibrium to shift to the right.

The change favours the formation of more product.

D. Adding more $\text{CaCO"_3"(s)}$

Adding more $\text{CaCO"_3"(s)}$ does not change the concentration (density) of $\text{CaCO"_3"(s)}$.

The change leads to no change in the position of equilibrium.

E. Adding $\text{LiNO"_3"(aq)}$

The net ionic equation is

$\text{Ca"^"2+""(aq)" + "CO"_2"(g)" + "2OH"^"-""(aq)" ⇌ "CaCO"_3"(s)" + "H"_2"O(l)}$

Neither ${\text{Li}}^{+}$ nor $\text{NO"_3^"-}$ participates in the reaction (they are spectator ions).

The change leads to no change in the position of equilibrium.