Lecture on Solving Equilibrium Concentration Problems

Introduction

• Solving for equilibrium concentrations involves using the equilibrium constant and initial concentrations.
• Problems may be specific to particular substances or the entire reaction.

Steps for Solving Equilibrium Problems

1. Set up a RICE Table
   • Initial conditions, Change, and Equilibrium.
2. Calculate Q
   • Determine the direction of the reaction.
3. Fill Change Row
   • Use values of x that match stoichiometry.
4. Complete Equilibrium Row
5. Substitute into Equilibrium Constant Expression
   • Solve for x.
6. Substitute x Back into Equilibrium Row
   • Find numerical values for equilibrium concentrations.

Example 1: Reaction of A in Equilibrium with 2B

• Given:
  • Equilibrium constant ( K = 0.33 )
  • Initial ( [A] = 1 ) M
• RICE Table Setup:
  • Reaction: ( A \rightleftharpoons 2B )
  • Initial: ( [A] = 1 ) M, ( [B] = 0 ) M
• Calculate Q:
  • ( Q < K ), reaction moves to the right.
• Change Row:
  • ( [A]: -x )
  • ( [B]: +2x )
• Equilibrium Row:
  • ( [A] = 1 - x )
  • ( [B] = 2x )
• Equilibrium Expression:
  • ( K = \frac{(2x)^2}{1-x} = 0.33 )
  • Solve: ( 4x^2 = 0.33(1-x) )
  • Rearrange: ( 4x^2 + 0.33x - 0.33 = 0 )
  • Quadratic Formula gives ( x = 0.2489 ) (accept) and ( x = -0.33 ) (reject)
• Final Concentrations:
  • ( [A] = 0.7511 ) M
  • ( [B] = 0.498 ) M

Example 2: Reaction of N2O4 with NO2

• Given:
  • ( K = 0.36 ) at 100°C
  • Initial ( [NO2] = 1 ) M, ( [N2O4] = 0.1 ) M
• RICE Table Setup:
  • Reaction: ( N2O4 \rightleftharpoons 2NO2 )
• Calculate Q:
  • ( Q = 10 ), reaction moves to the left.
• Change Row:
  • ( [N2O4]: +x )
  • ( [NO2]: -2x )
• Equilibrium Row:
  • ( [N2O4] = 0.1 + x )
  • ( [NO2] = 1 - 2x )
• Equilibrium Expression:
  • ( K = \frac{(1-2x)^2}{0.1+x} = 0.36 )
  • Expand and solve using quadratic formula: ( 4x^2 - 4.36x + 0.964 = 0 )
  • Solutions: ( x = 0.7817 ) (reject) and ( x = 0.3083 ) (accept)
• Final Concentrations:
  • To be completed by plugging x back into the equilibrium row.