Mercury and oxygen react to form mercury(II) oxide as shown below:

2 Hg(l) + O_{2}(g) ⇌ 2 HgO(s)

At a certain temperature in a 8.4 L reaction vessel that has reached equilibrium, the chemical components have the following composition:

Which of the following is true if the mass of all the chemical components double and the volume remains constant? (Hg: 48.8 g, O_{2}: 11.8 g, HgO: 37.2 g)

A. Q > K, shift left

B. Q > K, shift right

C. Q < K, shift left

D. Q < K, shift right

E. Q = K, no shift

We are asked to **calculate the Q and K values and determine the shift in the reaction. **

**The formula for Q is:**

$\overline{){\mathbf{Q}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{products}}{\mathbf{reactants}}}$

**The formula for K is:**

$\overline{){\mathbf{K}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{{\mathbf{products}}_{\mathbf{eq}}}{{\mathbf{reactants}}_{\mathbf{eq}}}}$

__Notice that K only differs from Q in a way that the concentration or pressures must be at equilibrium for K but not for Q. __

Depending on if Q is greater than or less than K, our reaction will shift to attain equilibrium by reaching the equilibrium constant K:

If **Q = K** → the reaction is at **equilibrium**

If **Q < K** → the reaction shifts in the **forward direction** to reach equilibrium

If **Q > K** → the reaction shifts in the **reverse direction** to reach equilibrium

__Take note that we don't include solids or liquids in the K and Q expressions. __

This means that in the problem:

2 Hg(l) + O_{2}(g) ⇌ 2 HgO(s)

**We are only concerned with O _{2}.**

**We go through the following steps to solve the problem: **

Step 1. Calculate the molar mass O_{2}

Step 2. Calculate the molarity O_{2} for both scenarios.

Step 3. Calculate Q and K.

Step 4. Compare Q and K.

The Reaction Quotient

The Reaction Quotient

The Reaction Quotient

The Reaction Quotient