# Question #763e3

Jun 8, 2015

The idea with dynamic equilibria is that any change in the amount/concentration of the products or reactants will be counteracted by a shift in equilibrium.

#### Explanation:

If something increases, the reaction that will consume that something will be favored. If something decreases, the reaction that will produce that something will be favored.

Let's say that you have an endothermic equilibrium reaction that looks like this

$\textcolor{red}{\text{heat}} + A + B r i g h t \le f t h a r p \infty n s C + D$ $\text{ } \textcolor{b l u e}{\left(1\right)}$

and an exothermic equilibrium reaction

$W + X r i g h t \le f t h a r p \infty n s Y + Z + \textcolor{red}{\text{heat}}$ $\text{ } \textcolor{b l u e}{\left(2\right)}$

Here are a couple of scenarios that you can encounter

• You add more $A$ to the first equilibrium

If you increase the concentration of one of the products, the equilibrium will shift in a way as to counteract this increase. This means that it will shift to the right, since the forward reaction will reduce the amount of reactants and increase the amount of products.

The favored reaction:

$\textcolor{red}{\text{heat}} + A + B \to C + D$

• You add more $C$ to the first equilibrium

This time, one of the products has a bigger concentration. Since the shift in equilibrium must counteract this change, a shift to the left will take place, since that would decrease the amount of products.

The favored reaction:

$C + D \to A + B + \textcolor{red}{\text{heat}}$

• You increase the temperature of the reaction in the second equilibrium

Notice that an exothermic reaction has temperature on the products' side. This means that when you increase temperature, you're essentially increasing the "concentration" of a product.

As a result, the equilibrium will shift to the left, since the reverse reaction will consume products, thus reduce the impact of the added heat, and form reactants.

The favored reaction:

$\textcolor{red}{\text{heat}} + Z + Y \to X + W$

• You increase the temperature of the reaction in the first equilibrium

This time, heat is on the reactants' side, which means that the equilibrium will have to shift to the right, since the forward reaction will consume the added heat.

Since heat is considered to be a reactant, the equilibrium will favor the consumption of the excess reactant, and thus produce more produts.

The favored reaction:

$\textcolor{red}{\text{heat}} + A + B \to C + D$

• You decrease the temperature of the first equilibrium

Removing heat from the reactants' side will cause the equilibrium to try and compensate for this change. As a result, the equilibrium will shift to the left, consuming products and producing heat.

Notice that the reverse reaction is actually exothermic, since now $C$ and $D$ react to produce $A$, $B$, and realease heat.

The favored reaction:

$C + D \to A + B + \textcolor{red}{\text{heat}}$

• You decrease the temperature of the second equilibrium

Removing heat from an exothermic reaction will cause the equilibrium to shift to the right. This will consume more of the reactants and release more heat.

The favored reaction:

$W + X \to Z + Y + \textcolor{red}{\text{heat}}$

• You add more $Z$ to the second equilibrium

Once again, you have an excess on the products' side. If the shift is to counteract this excess, the reverse reaction will have to be favored.

This implies that more products will be consumed and more reactants formed.

The favored reaction:

$\textcolor{red}{\text{heat}} + Z + Y \to W + X$