Reaction Quotient and Equilibrium Notes

Reaction Quotient (Q) and Equilibrium Constant (K)

• The equilibrium constant (Keq) can be represented as Kc (equilibrium constant with respect to concentration) or Kp (equilibrium constant with respect to pressure).
• If a system is not at equilibrium, the mass action expression does not equal the equilibrium constant.

Reaction Quotient (Q)

• The reaction quotient (Q) is used to designate that the system is not at equilibrium.
• As the reaction proceeds ($Q$ approaches $K$), the system gets closer to equilibrium.
• When $Q = K$, the system is at equilibrium.

Scenarios and Shifts

• If $Q > K$, the reaction shifts to make more reactants (shifts to the left).
  • The ratio of products to reactants is too high, so the denominator (reactants) needs to increase.
• If $K > Q$, the reaction shifts to make more products (shifts to the right).
• If $K = Q$, the system is already at equilibrium.

Example

Consider a system in equilibrium with the number of moles of reacting species given in a flask at a specific temperature.

1. Write the mass action expression for Q:
   $Q = \frac{[SO<em>2][Cl</em>2]}{[SO<em>2Cl</em>2]}$

2. Convert moles to concentration (Molarity):
   $C = \frac{n}{V}$, where:

   • $C$ is concentration (Molarity).

   • $n$ is the number of moles.

   • $V$ is the volume in liters.

   • Given:

     • $SO<em>2Cl</em>2$ = 0.5 moles in 5 L
     • $SO_2$ = 0.035 moles in 5 L
     • $Cl_2$ = 0.08 moles in 5 L

   • Calculations:

     • $[SO<em>2Cl</em>2] = \frac{0.5 \text{ moles}}{5 \text{ L}} = 0.1 \text{ M}$
     • $[SO_2] = \frac{0.035 \text{ moles}}{5 \text{ L}} = 0.007 \text{ M}$
     • $[Cl_2] = \frac{0.08 \text{ moles}}{5 \text{ L}} = 0.016 \text{ M}$

3. Substitute the concentration values into the Q expression:

   $Q = \frac{[0.007][0.016]}{[0.1]} = 0.0011$

4. Compare Q to K:

   • Given: $K = 0.078$
   • Calculated: $Q = 0.0011$
   • Since $K > Q$, the reaction needs to proceed forward to make more products until equilibrium is established.