Calculating K
Related Examples and Practice Problems
Additional Worked Out Examples/ Practice
Basic Calculation of Kc: Use the balanced chemical equation and equilibrium concentrations of reactants and products to calculate Kc
Separation techniques: Selected and explaining limitation of appropriate separation
Relating Properties to Composition: Predicting classification based on descriptive properties
Topic Summary & Highlights
and Help Videos
Core Concept
The equilibrium constant quantifies the ratio of the concentrations (or partial pressures) of products to reactants at equilibrium for a given chemical reaction.
There are two forms of equilibrium constants:
Kc: Based on molar concentrations (mol/L).
Kp: Based on partial pressures (atm).
For a generic reaction: aA + bB ⇌ cC + dD
The equilibrium constant is expressed as:
$K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$
$K_p = \frac{(P_{\text{C}})^c (P_{\text{D}})^d}{(P_{\text{A}})^a (P_{\text{B}})^b}$
Practice Tips
Check Stoichiometry: Ensure coefficients in the balanced equation are correctly applied in the K expression.
Keep Units Consistent: Use concentrations for Kc and partial pressures for Kp.
Temperature Matters: If solving for Kp using Kc, remember to convert temperature to Kelvin.
Don’t Include Solids or Liquids: Pure solids and liquids do not appear in the K expression because their concentrations are constant.
Core Concept
The equilibrium constant quantifies the ratio of the concentrations (or partial pressures) of products to reactants at equilibrium for a given chemical reaction.
There are two forms of equilibrium constants:
Kc: Based on molar concentrations (mol/L).
Kp: Based on partial pressures (atm).
For a generic reaction: aA + bB ⇌ cC + dD
The equilibrium constant is expressed as:
$K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$
$K_p = \frac{(P_{\text{C}})^c (P_{\text{D}})^d}{(P_{\text{A}})^a (P_{\text{B}})^b}$
Example Problem: Calculate Kc Using Equilibrium Concentrations
For the reaction: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$
Given:
$[\text{H}_2] = 0.10 \, \text{M}$
$[\text{I}_2] = 0.10 \, \text{M}$
$[\text{HI}] = 0.80 \, \text{M}$
Solution:
Write the expression for Kc: $K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}$
Substitute values: $K_c = \frac{(0.80)^2}{(0.10)(0.10)} = \frac{0.64}{0.01} = 64.0$