How To Calculate Equilibrium Partial Pressure

In the realm of chemistry, understanding equilibrium is crucial for predicting the behavior of reactions. One aspect of equilibrium is the partial pressure of gases in a system. Calculating equilibrium partial pressures allows us to determine the composition of a gaseous mixture at equilibrium and understand the extent to which a reaction proceeds. This article delves into the methods for calculating equilibrium partial pressures, providing a comprehensive guide with examples and explanations.

Understanding Equilibrium Partial Pressure

Before diving into calculations, it's important to understand the concept of partial pressure and its relation to equilibrium.

Partial pressure is the pressure exerted by a single gas in a mixture of gases. According to Dalton's Law of Partial Pressures, the total pressure of a mixture of gases is equal to the sum of the partial pressures of each individual gas. Mathematically, this is expressed as:

P_total = P₁ + P₂ + P₃ + ... + P_n

where P_total is the total pressure of the mixture, and P₁, P₂, P₃, ..., P_n are the partial pressures of each gas component.

Equilibrium is the state where the rates of the forward and reverse reactions are equal, and there is no net change in the concentrations of reactants and products. For gaseous reactions, equilibrium can be expressed in terms of partial pressures. The equilibrium constant K_p is defined as the ratio of the partial pressures of products to reactants, each raised to the power of their stoichiometric coefficients.

For a general reversible reaction:

aA(g) + bB(g) ⇌ cC(g) + dD(g)

The equilibrium constant K_p is given by:

K_p = (P_C^c * P_D^d) / (P_A^a * P_B^b)

where P_A, P_B, P_C, and P_D are the equilibrium partial pressures of gases A, B, C, and D, respectively, and a, b, c, and d are their stoichiometric coefficients.

Steps to Calculate Equilibrium Partial Pressures

Calculating equilibrium partial pressures typically involves the following steps:

1. Write the balanced chemical equation: Ensure the reaction is properly balanced to determine the correct stoichiometric coefficients.
2. Set up an ICE table: The ICE table (Initial, Change, Equilibrium) is a helpful tool for organizing the information and tracking the changes in partial pressures as the reaction reaches equilibrium.
3. Express equilibrium partial pressures in terms of 'x': Define 'x' as the change in partial pressure of a reactant or product based on the stoichiometry of the reaction. Express the equilibrium partial pressures of all gases in terms of 'x'.
4. Substitute into the K_p expression: Plug the equilibrium partial pressure expressions into the K_p expression.
5. Solve for 'x': Solve the resulting equation for 'x'. This may involve using the quadratic formula or making simplifying assumptions if 'x' is small compared to the initial partial pressures.
6. Calculate equilibrium partial pressures: Substitute the value of 'x' back into the expressions for the equilibrium partial pressures to find the partial pressure of each gas at equilibrium.
7. Verify the answer: Check if the calculated equilibrium partial pressures are reasonable and make sense in the context of the problem.

Example 1: Simple Equilibrium Calculation

Consider the following reaction at a certain temperature:

N₂O₄(g) ⇌ 2NO₂(g)

The equilibrium constant K_p is 0.14. Initially, a flask contains only N₂O₄ at a partial pressure of 0.50 atm. Calculate the equilibrium partial pressures of N₂O₄ and NO₂.

Solution:

1. Balanced Equation: N₂O₄(g) ⇌ 2NO₂(g)

2. ICE Table:

   	N<sub>2</sub>O<sub>4</sub>	2NO<sub>2</sub>
   Initial (I)	0.50 atm	0 atm
   Change (C)	-x	+2x
   Equilibrium (E)	0.50 - x	2x

3. Equilibrium Partial Pressures in Terms of 'x':

   • P_N2O4 = 0.50 - x
   • P_NO2 = 2x

4. Substitute into the K_p expression:

   K_p = (P_NO2)² / P_N2O4 0. 14 = (2x)² / (0.50 - x)

5. Solve for 'x':

   0. 14(0.50 - x) = 4x²
   1. 07 - 0.14x = 4x²
   2. x² + 0.14x - 0.07 = 0

   Using the quadratic formula:

   x = (-b ± √(b² - 4ac)) / 2a x = (-0.14 ± √((0.14)² - 4(4)(-0.07))) / (2 * 4) x = (-0.14 ± √(0.0196 + 1.12)) / 8 x = (-0.14 ± √1.1396) / 8 x = (-0.14 ± 1.0675) / 8

   We have two possible values for x:

   • x = (-0.14 + 1.0675) / 8 = 0.116
   • x = (-0.14 - 1.0675) / 8 = -0.151 (This value is not physically meaningful since partial pressures cannot be negative)

   Therefore, x = 0.116

6. Calculate Equilibrium Partial Pressures:

   • P_N2O4 = 0.50 - x = 0.50 - 0.116 = 0.384 atm
   • P_NO2 = 2x = 2 * 0.116 = 0.232 atm

7. Verify the Answer:

   Check if the calculated values are reasonable. Also, check if the K_p value matches:

   K_p = (0.232)² / 0.384 = 0.053824 / 0.384 = 0.14

The equilibrium partial pressures are: P_N2O4 = 0.384 atm and P_NO2 = 0.232 atm.

Example 2: Equilibrium with Initial Partial Pressures of Both Reactants and Products

Consider the following reaction:

H₂(g) + I₂(g) ⇌ 2HI(g)

At 448°C, the equilibrium constant K_p is 50.0. Initially, a mixture contains H₂ at a partial pressure of 0.50 atm, I₂ at a partial pressure of 0.50 atm, and HI at a partial pressure of 0.50 atm. Calculate the equilibrium partial pressures of all gases.

Solution:

1. Balanced Equation: H₂(g) + I₂(g) ⇌ 2HI(g)

2. ICE Table:

   	H<sub>2</sub>	I<sub>2</sub>	2HI
   Initial (I)	0.50 atm	0.50 atm	0.50 atm
   Change (C)	-x	-x	+2x
   Equilibrium (E)	0.50 - x	0.50 - x	0.50 + 2x

3. Equilibrium Partial Pressures in Terms of 'x':

   • P_H2 = 0.50 - x
   • P_I2 = 0.50 - x
   • P_HI = 0.50 + 2x

4. Substitute into the K_p expression:

   K_p = (P_HI)² / (P_H2 * P_I2) 50. 0 = (0.50 + 2x)² / ((0.50 - x)(0.50 - x)) 51. 0 = (0.50 + 2x)² / (0.50 - x)²

5. Solve for 'x':

   Take the square root of both sides:

   √50 = (0.50 + 2x) / (0.50 - x) 72 = (0.50 + 2x) / (0.50 - x) 73. 07(0.50 - x) = 0.50 + 2x 74. 535 - 7.07x = 0.50 + 2x 75. 035 = 9.07x x = 3.035 / 9.07 = 0.335

6. Calculate Equilibrium Partial Pressures:

   • P_H2 = 0.50 - x = 0.50 - 0.335 = 0.165 atm
   • P_I2 = 0.50 - x = 0.50 - 0.335 = 0.165 atm
   • P_HI = 0.50 + 2x = 0.50 + 2(0.335) = 0.50 + 0.67 = 1.17 atm

7. Verify the Answer:

   K_p = (1.17)² / (0.165 * 0.165) = 1.3689 / 0.027225 = 50.3

The equilibrium partial pressures are: P_H2 = 0.165 atm, P_I2 = 0.165 atm, and P_HI = 1.17 atm.

Example 3: Using the Ideal Gas Law

Sometimes, the initial conditions are given in terms of moles and volume. In such cases, the ideal gas law (PV = nRT) can be used to calculate the initial partial pressures.

Consider the following reaction:

CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)

A 10.0 L vessel at 700 K initially contains 0.10 mol of CO and 0.10 mol of H₂O. At equilibrium, it is found that the vessel contains 0.065 mol of CO₂. Calculate the equilibrium constant K_p and the equilibrium partial pressures of all gases.

Solution:

1. Balanced Equation: CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)

2. Calculate Initial Partial Pressures:

   Using the Ideal Gas Law: PV = nRT P = nRT/V R = 0.0821 L atm / (mol K)

   • P_CO (initial) = (0.10 mol * 0.0821 L atm / (mol K) * 700 K) / 10.0 L = 0.5747 atm
   • P_H2O (initial) = (0.10 mol * 0.0821 L atm / (mol K) * 700 K) / 10.0 L = 0.5747 atm
   • P_CO2 (initial) = 0 atm
   • P_H2 (initial) = 0 atm

3. Calculate Change in Moles and Partial Pressures:

   At equilibrium, 0.065 mol of CO₂ is present. Δn_CO2 = 0.065 mol ΔP_CO2 = (0.065 mol * 0.0821 L atm / (mol K) * 700 K) / 10.0 L = 0.373 atm

4. ICE Table:

   	CO	H<sub>2</sub>O	CO<sub>2</sub>	H<sub>2</sub>
   Initial (I)	0.5747	0.5747	0	0
   Change (C)	-0.373	-0.373	+0.373	+0.373
   Equilibrium (E)	0.2017	0.2017	0.373	0.373

5. Equilibrium Partial Pressures:

   • P_CO = 0.5747 - 0.373 = 0.2017 atm
   • P_H2O = 0.5747 - 0.373 = 0.2017 atm
   • P_CO2 = 0.373 atm
   • P_H2 = 0.373 atm

6. Calculate K_p:

   K_p = (P_CO2 * P_H2) / (P_CO * P_H2O) K_p = (0.373 * 0.373) / (0.2017 * 0.2017) K_p = 0.139129 / 0.040683 = 3.42

7. Verify the Answer:

   The K_p value and the calculated partial pressures are reasonable.

The equilibrium partial pressures are: P_CO = 0.2017 atm, P_H2O = 0.2017 atm, P_CO2 = 0.373 atm, and P_H2 = 0.373 atm. The equilibrium constant K_p is 3.42.

Simplifying Assumptions

In some cases, solving for 'x' can be complicated, especially if the equilibrium constant K_p is very small. If K_p is small enough, we can make the assumption that 'x' is negligible compared to the initial partial pressures. This simplifies the algebra and makes the problem easier to solve.

For example, if the initial partial pressure of a reactant is 1.0 atm and K_p is very small, we can assume that (1.0 - x) ≈ 1.0. However, it's important to check the validity of this assumption after solving for 'x'. A common rule of thumb is that if 'x' is less than 5% of the initial partial pressure, the assumption is valid. If 'x' is greater than 5%, the quadratic formula or another method must be used to solve for 'x' accurately.

Factors Affecting Equilibrium Partial Pressures

Several factors can affect the equilibrium partial pressures in a gaseous reaction:

• Temperature: According to Le Chatelier's principle, changing the temperature will shift the equilibrium position to favor either the forward or reverse reaction, depending on whether the reaction is endothermic or exothermic.
• Pressure: Changing the total pressure of the system will affect the equilibrium position if the number of moles of gas is different on the reactant and product sides.
• Initial Partial Pressures: The initial partial pressures of reactants and products will influence the direction in which the reaction shifts to reach equilibrium.
• Presence of Inert Gases: Adding an inert gas to the system at constant volume will not affect the equilibrium partial pressures because it does not participate in the reaction.

Common Mistakes to Avoid

• Forgetting to Balance the Chemical Equation: Always start with a balanced chemical equation to ensure the correct stoichiometric coefficients are used in the K_p expression and ICE table.
• Incorrectly Setting up the ICE Table: Pay attention to the stoichiometry of the reaction when determining the changes in partial pressures in the ICE table.
• Using Incorrect Units: Make sure to use consistent units for partial pressures (e.g., atm, kPa) and the equilibrium constant K_p.
• Not Checking the Validity of Simplifying Assumptions: If you make a simplifying assumption, verify that 'x' is indeed small enough compared to the initial partial pressures.
• Algebra Errors: Be careful with algebraic manipulations when solving for 'x', especially when using the quadratic formula.

Conclusion

Calculating equilibrium partial pressures is a fundamental skill in chemistry that allows us to understand and predict the behavior of gaseous reactions. By following the steps outlined in this article, including writing the balanced chemical equation, setting up an ICE table, expressing equilibrium partial pressures in terms of 'x', substituting into the K_p expression, and solving for 'x', you can confidently calculate the equilibrium partial pressures of gases in a variety of systems. Remember to verify your answers and be aware of potential simplifying assumptions and common mistakes. With practice, you'll become proficient in this important aspect of chemical equilibrium.