For gas-phase reactions in equilibrium, we define the equilibrium constant \( K_p \) using the partial pressures of the reactants and products at equilibrium. This lesson covers how to write \( K_p \) expressions, perform calculations, deduce units, and understand why temperature is the only factor that alters the value of the constant.
🔑 Key Principle
Only species in the gas phase are included in the expression for \( K_p \). Any solid or liquid reactants or products are omitted. This is because their concentration or effective pressure remains constant during the reaction.
Writing Kp Expressions
For a general gas-phase reaction:
\( aA\text{(g)} + bB\text{(g)} \rightleftharpoons cC\text{(g)} + dD\text{(g)} \)
The equilibrium constant \( K_p \) expression is written as:
Where \( p_X \) represents the partial pressure of gas \( X \) at equilibrium.
The constant that relates the partial pressures of reactants and products at equilibrium for a gas-phase reaction at a specific temperature.
Omit Solids and Liquids (Heterogeneous Equilibria)
If a reaction involves solids or liquids alongside gases, omit the non-gaseous species from the \( K_p \) expression entirely.
For example, in the thermal decomposition of calcium carbonate:
\( \text{CaCO}_3\text{(s)} \rightleftharpoons \text{CaO(s)} + \text{CO}_2\text{(g)} \)
The expression for \( K_p \) is simply:
\( K_p = p_{\text{CO}_2} \)
Never use square brackets in a \( K_p \) expression. Square brackets, e.g., \( [\text{CO}_2] \), denote concentrations and are used for \( K_c \). If you write square brackets in a \( K_p \) question on an exam, you will lose the mark for the expression and potentially all subsequent calculation marks.
Calculating Kp Units
Units for \( K_p \) are derived by substituting pressure units (e.g. \( \text{kPa} \), \( \text{atm} \), or \( \text{Pa} \)) into the \( K_p \) expression and simplifying. Unlike \( K_c \), the units depend directly on the units given for pressure in the question.
Step 1: Write the \( K_p \) expression:
\( K_p = \frac{(p_{\text{SO}_3})^2}{(p_{\text{SO}_2})^2 \times p_{\text{O}_2}} \)
Step 2: Substitute kPa into the expression:
\( \text{Units of } K_p = \frac{(\text{kPa})^2}{(\text{kPa})^2 \times \text{kPa}} \)
Step 3: Cancel units:
\( \text{Units of } K_p = \frac{1}{\text{kPa}} = \text{kPa}^{-1} \)
Worked Example 2: Complete Kp Calculation
Step 1: Set up the ICE table:
| Species | \( \text{N}_2\text{(g)} \) | \( 3\text{H}_2\text{(g)} \) | \( 2\text{NH}_3\text{(g)} \) |
|---|---|---|---|
| Initial moles | 1.00 | 3.00 | 0.00 |
| Change in moles | -0.20 | -0.60 | +0.40 |
| Equilibrium moles | 0.80 | 2.40 | 0.40 |
Note: Since 0.40 mol of \( \text{NH}_3 \) was formed, nitrogen decreased by 0.20 mol (1:2 ratio) and hydrogen decreased by 0.60 mol (3:2 ratio).
Step 2: Calculate total moles:
\( n_{\text{total}} = 0.80 + 2.40 + 0.40 = 3.60\text{ mol} \)
Step 3: Calculate mole fractions (\( \chi \)):
- \( \chi_{\text{N}_2} = \frac{0.80}{3.60} = 0.2222 \)
- \( \chi_{\text{H}_2} = \frac{2.40}{3.60} = 0.6667 \)
- \( \chi_{\text{NH}_3} = \frac{0.40}{3.60} = 0.1111 \)
Step 4: Calculate partial pressures (\( p = \chi \times P_{\text{total}} \)):
- \( p_{\text{N}_2} = 0.2222 \times 200 = 44.44\text{ kPa} \)
- \( p_{\text{H}_2} = 0.6667 \times 200 = 133.34\text{ kPa} \)
- \( p_{\text{NH}_3} = 0.1111 \times 200 = 22.22\text{ kPa} \)
Step 5: Substitute partial pressures into the Kp expression and calculate:
\( K_p = \frac{(p_{\text{NH}_3})^2}{p_{\text{N}_2} \times (p_{\text{H}_2})^3} \)
\( K_p = \frac{(22.22)^2}{44.44 \times (133.34)^3} \)
\( K_p = \frac{493.73}{44.44 \times 2370603.95} = \frac{493.73}{105349639.5} = 4.69 \times 10^{-6} \)
Step 6: Determine units:
\( \text{Units} = \frac{(\text{kPa})^2}{\text{kPa} \times (\text{kPa})^3} = \frac{1}{\text{kPa}^2} = \text{kPa}^{-2} \)
Final Answer: \( K_p = 4.69 \times 10^{-6}\text{ kPa}^{-2} \) (to 3 significant figures).
Effect of Changing Conditions on Kp
The position of equilibrium changes when conditions are altered, but the equilibrium constant \( K_p \) itself remains constant for all changes except temperature.
Effect of Temperature
Temperature is the only factor that changes the value of \( K_p \):
- For an exothermic reaction (\( \Delta H < 0 \)): Increasing the temperature shifts the equilibrium in the endothermic direction (to the left). This reduces the partial pressure of products and increases the partial pressure of reactants, so the value of \( K_p \) decreases.
- For an endothermic reaction (\( \Delta H > 0 \)): Increasing the temperature shifts the equilibrium in the endothermic direction (to the right). This increases the partial pressure of products and decreases the partial pressure of reactants, so the value of \( K_p \) increases.
Effect of Pressure and Concentration
Changing the total pressure or concentrations of the species alters the position of equilibrium, but does not affect the value of \( K_p \). According to the expression:
\( K_p = \frac{\text{products}}{\text{reactants}} \)
If the pressure is increased, the partial pressures of all species change. The system shifts to restore the ratio, ensuring the calculated value of \( K_p \) remains exactly the same.
Effect of a Catalyst
A catalyst increases the rate of both the forward and reverse reactions by the same factor. Equilibrium is reached faster, but the position of equilibrium and the value of \( K_p \) are completely unchanged.
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