The Equilibrium Constant Kp Is 158- Calculations

What Kp Actually Is

Kp is the equilibrium constant expressed in terms of partial pressures. When a reaction involves gases, concentrations won't cut it—you need pressure. That's where Kp comes in.

Unlike Kc (which uses molarity), Kp uses the partial pressures of gaseous reactants and products at equilibrium. Simple. Direct. No hidden meanings.

Kp vs Kc: The Relationship

These two constants are related through a single equation:

Kp = Kc(RT)^Δn

Where:

• R = 0.0821 L·atm/(mol·K) or 8.314 J/(mol·K)
• T = temperature in Kelvin
• Δn = moles of gaseous products minus moles of gaseous reactants

If Δn = 0, then Kp = Kc. That's the only time you can interchange them.

Writing Kp Expressions

For a general reaction:

aA(g) + bB(g) ⇌ cC(g) + dD(g)

The Kp expression is:

Kp = (Pc)^c × (Pd)^d / (Pa)^a × (Pb)^b

Same logic as Kc—just swap concentrations for partial pressures. Only include gaseous species. Solids and liquids get excluded because their activities equal 1.

Example

For the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Kp = (P_NH₃)² / (P_N₂) × (P_H₂)³

Notice the coefficients become exponents. That's the rule.

How to Solve Kp Problems

Problem Type 1: Finding Kp from Partial Pressures

Given: At equilibrium in a 2.0 L container at 400 K:

• P_N₂ = 0.40 atm
• P_H₂ = 0.60 atm
• P_NH₃ = 0.20 atm

Find: Kp for N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Solution:

Just plug into the expression:

Kp = (0.20)² / (0.40) × (0.60)³

Kp = 0.04 / 0.40 × 0.216

Kp = 0.04 / 0.0864

Kp = 0.463

Problem Type 2: Finding Equilibrium Partial Pressures

Given: Kp = 158 for: 2NO₂(g) ⇌ 2NO(g) + O₂(g)

Initial pressures: P_NO₂ = 1.0 atm, P_NO = 0 atm, P_O₂ = 0 atm

Find: Equilibrium partial pressures

Solution:

Set up an ICE table (Initial, Change, Equilibrium):

Write the Kp expression:

158 = (2x)² × (x) / (1.0-2x)²

158 = 4x³ / (1.0-2x)²

Solve for x. This requires either quadratic formula or approximation. Using approximation (assuming x is small):

158 ≈ 4x³

x³ ≈ 39.5

x ≈ 3.4 atm

But wait—1.0 - 2(3.4) = -5.8. That doesn't work. The assumption fails. Use quadratic formula properly:

158(1.0 - 4x + 4x²) = 4x³

158 - 632x + 632x² = 4x³

4x³ - 632x² + 632x - 158 = 0

Solving gives x ≈ 0.29 atm

Equilibrium pressures:

• P_NO₂ = 1.0 - 2(0.29) = 0.42 atm
• P_NO = 2(0.29) = 0.58 atm
• P_O₂ = 0.29 atm

Converting Between Kp and Kc

When you need to switch between these constants, use this formula:

Kp = Kc × (RT)^Δn

Example: Kc = 0.045 at 500 K for: CO(g) + 2H₂(g) ⇌ CH₃OH(g)

Δn = 1 - 3 = -2

Kp = 0.045 × (0.0821 × 500)^(-2)

Kp = 0.045 × (41.05)^(-2)

Kp = 0.045 / 1685

Kp = 2.67 × 10⁻⁵

Common Mistakes That Kill Your Answers

• Forgetting to convert °C to Kelvin. Always add 273 to Celsius temperatures.
• Including liquids and solids in the expression. They don't belong. Their activity is 1.
• Getting Δn wrong. Count only gaseous species. Spectator ions don't count.
• Forgetting units. Kp has no units when using atmospheres, but check your textbook's convention.
• Not checking if the approximation is valid. If x is more than 5% of initial values, the approximation fails.

Quick Reference Table

Practical How-To: Solving Any Kp Problem in 5 Steps

1. Write the balanced equation. This is non-negotiable. You can't solve what you haven't balanced.
2. Identify gaseous species only. Skip liquids and solids—they don't appear in the expression.
3. Write the Kp expression. Products over reactants, coefficients as exponents.
4. Build an ICE table if needed. Track initial pressures, changes, and equilibrium values.
5. Substitute and solve. Plug equilibrium values into Kp expression. Use algebra to find unknowns.

The Bottom Line

Kp calculations follow the same logic as Kc—the only difference is you're working with partial pressures instead of concentrations. Master the expression, watch your Δn, and convert units properly. That's it.

When you see Kp = 158 in a problem, don't panic. Set up your ICE table, substitute into the expression, and solve for x. The math is straightforward once you know the steps.