Comprehensive Study Notes: Equilibrium Constant Kp for Homogeneous Systems (9620 A2 Chemistry)

Hello future chemist! This topic, Equilibrium Constant Kp, is where we apply everything you learned about dynamic equilibrium and Le Chatelier's principle, but specifically for reactions involving gases. It's super important for industrial processes like the Haber process, where controlling gases and pressure is key to making products efficiently. Don't worry if the calculations look intimidating—they simply require a methodical, step-by-step approach!


1. Understanding Pressure in Gas Mixtures

The constant \(K_p\) uses the partial pressures of the gases, not their concentrations (which we used for \(K_c\)). Since all reactions we study here are homogeneous (meaning all reactants and products are in the gas phase), pressure is the ideal way to measure "how much" of each substance is present.

1.1 Partial Pressure and Total Pressure

Imagine a container filled with a mixture of gases (A, B, and C). The total pressure inside the container is simply the sum of the pressures exerted by each individual gas. This individual pressure is called the partial pressure.

\(P_{\text{Total}} = P_A + P_B + P_C + \dots\)

The partial pressure of a gas is directly related to how many moles of that gas are present compared to the total number of moles of gas.

1.2 The Mole Fraction (\(x\))

The mole fraction (\(x\)) of a gas tells you the proportion of that gas in the total mixture. It's calculated by comparing the moles of the specific gas to the total moles of all gases present at equilibrium.

\(x_{\text{gas A}} = \frac{\text{Number of moles of gas A}}{\text{Total number of moles of all gases in the mixture}}\)

1.3 Deriving Partial Pressure (\(P\))

The partial pressure of a gas is found by multiplying its mole fraction by the total pressure of the system:

$$P_{\text{gas A}} = x_{\text{gas A}} \times P_{\text{Total}}$$

Example: If gas A makes up 25% of the total moles (mole fraction = 0.25) and the total pressure is 100 kPa, then the partial pressure of A is \(0.25 \times 100 \text{ kPa} = 25 \text{ kPa}\).


Quick Review: Partial Pressure
  • Partial pressure is the pressure one gas exerts.
  • It requires two things: Mole Fraction and Total Pressure.
  • \(P_{\text{gas}} = x_{\text{gas}} \times P_{\text{Total}}\).

2. Constructing the \(K_p\) Expression

The Equilibrium Constant \(K_p\) is the mathematical expression derived from the balanced chemical equation, using the partial pressures of the gaseous reactants and products.

2.1 The General Form

Consider a general reversible reaction where all components (A, B, C, D) are gases:

\(aA_{(\text{g})} + bB_{(\text{g})} \rightleftharpoons cC_{(\text{g})} + dD_{(\text{g})}\)

The expression for \(K_p\) is:

$$K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}$$

  • The partial pressures of the products are on the top (numerator).
  • The partial pressures of the reactants are on the bottom (denominator).
  • Each partial pressure is raised to the power of its stoichiometric coefficient (the balancing number) from the equation.

Important Rule: Unlike \(K_c\), we DO NOT use square brackets [ ] for partial pressure. We use \(P\) with the substance's symbol as a subscript (e.g., \(P_{\text{NH}_3}\)).

Did you know? If a component is a liquid or solid (heterogeneous equilibrium), it is left out of the \(K_p\) expression because the "concentration" or "pressure contribution" of a pure solid or liquid is considered constant. Since we are focusing on homogeneous systems here, everything is included!

2.2 Determining Units of \(K_p\)

Since partial pressure is typically measured in kilopascals (kPa) or atmospheres (atm), \(K_p\) often has units. To find the units, look at the powers in the expression:

$$ \text{Units of } K_p = \frac{(\text{Unit of Pressure})^{\text{moles of product gas}}}{(\text{Unit of Pressure})^{\text{moles of reactant gas}}} $$

If the total moles of gas on the product side equals the total moles of gas on the reactant side, then \(K_p\) is dimensionless (it has no units).

Example: For the Haber process:

\(\text{N}_{2(\text{g})} + 3\text{H}_{2(\text{g})} \rightleftharpoons 2\text{NH}_{3(\text{g})}\)

The \(K_p\) expression is: \(K_p = \frac{(P_{\text{NH}_3})^2}{(P_{\text{N}_2}) (P_{\text{H}_2})^3}\)

If the pressure unit is atm:

$$ \text{Units} = \frac{(\text{atm})^2}{(\text{atm}) (\text{atm})^3} = \frac{\text{atm}^2}{\text{atm}^4} = \text{atm}^{-2} $$


3. Step-by-Step \(K_p\) Calculations

Calculations involving \(K_p\) often follow a structure similar to \(K_c\) calculations, but require the intermediate step of finding mole fractions and partial pressures.

Scenario: You are given initial moles and information about the change, along with the total pressure (\(P_{\text{Total}}\)).

Step-by-Step Guide to Calculating \(K_p\)

Step 1: Calculate Equilibrium Moles (The ICE table)
Use the initial amounts, the change, and the reaction stoichiometry to determine the moles of each gas present at equilibrium.

Step 2: Calculate Total Moles
Sum up the equilibrium moles of all gaseous components.

$$\text{Total Moles} = n_A + n_B + n_C + n_D$$

Step 3: Calculate Mole Fractions (\(x\))
Calculate the mole fraction for each gas at equilibrium:

$$x_{\text{gas}} = \frac{n_{\text{gas}}}{\text{Total Moles}}$$

Step 4: Calculate Partial Pressures (\(P\))
Calculate the partial pressure for each gas using the given total pressure:

$$P_{\text{gas}} = x_{\text{gas}} \times P_{\text{Total}}$$

Step 5: Substitute into the \(K_p\) Expression
Plug the calculated partial pressure values into the \(K_p\) equation you constructed in Section 2, and calculate the final value (don't forget the units!).

Don't worry if this seems tricky at first. Practice makes perfect! Most calculation mistakes happen in Step 1 (getting the equilibrium moles right) or Step 4 (forgetting to use the \(P_{\text{Total}}\)).


Common Mistake Alert!

When calculating \(K_p\), you must use partial pressures. You cannot skip straight from equilibrium moles to the \(K_p\) expression unless the question explicitly states that total pressure is 1 atm or 1 kPa, *and* that the change in moles is zero (which is rare).


4. Factors Affecting \(K_p\) and Equilibrium Position

We often use the principles established in the AS course (Le Chatelier's Principle) to predict how changes affect equilibrium, but here we look specifically at how these changes influence the calculated value of \(K_p\).

4.1 Effect of Temperature on \(K_p\)

Temperature is the ONLY factor that changes the VALUE of the equilibrium constant, \(K_p\).

  • If T increases: The equilibrium shifts in the direction that absorbs heat (the endothermic direction) to counteract the change (Le Chatelier's).
    • If the forward reaction is endothermic, increasing T shifts equilibrium right (more products). \(K_p\) increases.
    • If the forward reaction is exothermic, increasing T shifts equilibrium left (fewer products). \(K_p\) decreases.
  • If T decreases: The equilibrium shifts in the direction that releases heat (the exothermic direction).
    • If the forward reaction is exothermic, decreasing T shifts equilibrium right. \(K_p\) increases.

Memory Aid: If the shift increases the proportion of products, \(K_p\) increases. If the shift decreases the proportion of products, \(K_p\) decreases.

4.2 Effect of Pressure on Equilibrium Position (but NOT \(K_p\) Value)

A change in the total pressure of the system only affects the position of equilibrium, not the value of \(K_p\) (as long as the temperature is kept constant).

  • If Total Pressure increases: Equilibrium shifts to the side with the fewer total moles of gas. This reduces the number of particles, relieving the pressure stress.
  • If Total Pressure decreases: Equilibrium shifts to the side with the greater total moles of gas.

Why does \(K_p\) remain constant when pressure changes?

When the total pressure changes, the partial pressures of all components change proportionately. When these new partial pressures are substituted back into the \(K_p\) expression, the proportionality change cancels out, resulting in the same \(K_p\) value. The system simply adjusts the mole fractions to maintain the constant ratio defined by \(K_p\).

4.3 Effect of Adding a Catalyst

A catalyst increases the rate of both the forward and reverse reactions equally.

  • A catalyst does not affect the position of equilibrium.
  • A catalyst does not affect the value of the equilibrium constant (\(K_p\)).
  • It simply allows the system to reach equilibrium faster.

4.4 Industrial Implications: The Compromise

For industrial processes (like the Haber process for ammonia), chemists often have to use a compromise temperature and pressure.

  • Pressure: High pressure favors the side with fewer moles (often the product). High pressure is expensive and dangerous to maintain, so a moderate or high pressure is used to maximize yield while remaining economical.
  • Temperature: If the reaction is exothermic (like the Haber process), a low temperature gives a high equilibrium yield (large \(K_p\)). However, a low temperature means a very slow rate. Therefore, a high enough temperature must be chosen to ensure the reaction proceeds quickly (economical rate), even if it means sacrificing some yield (smaller \(K_p\)).

Key Takeaways for \(K_p\)
  • \(K_p\) is defined using the partial pressures of gases.
  • $$P_{\text{gas}} = \frac{n_{\text{gas}}}{n_{\text{Total}}} \times P_{\text{Total}}$$
  • Only a change in Temperature changes the value of \(K_p\).
  • Changes in Pressure or addition of a Catalyst only affect the rate of attainment or the position of equilibrium, not the value of \(K_p\).