Calculating Kp from Kc: Example Problems

Explore practical examples of calculating Kp from Kc in chemical equilibrium.
By Jamie

Understanding Kp and Kc

In chemical equilibrium, Kp and Kc are essential concepts that describe the relationship between the concentrations of reactants and products in a reaction at equilibrium. Kp refers to the equilibrium constant in terms of partial pressures, while Kc refers to the equilibrium constant in terms of molar concentrations. The two constants are related through the ideal gas law and the change in the number of moles of gas in a reaction. This guide presents three diverse, practical examples of calculating Kp from Kc.

Example 1: Calculating Kp for a Simple Reaction

In the synthesis of ammonia, the reaction is given by:

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

For this reaction, the equilibrium constant Kc is found to be 0.5 at a certain temperature. To find Kp, we can use the formula:

\[ K_p = K_c(RT)^{\Delta n} \]

Where:

  • R is the ideal gas constant (0.0821 L·atm/(K·mol))
  • T is the temperature in Kelvin
  • \( \Delta n \) is the change in the number of moles of gas, calculated as the number of moles of products minus the number of moles of reactants.

In this reaction:

  • Moles of products = 2 (from 2NH3)
  • Moles of reactants = 1 + 3 = 4 (from N2 and 3H2)
  • \( \Delta n = 2 - 4 = -2 \)

Assuming the temperature is 298 K:

\[ K_p = 0.5 \times (0.0821 \times 298)^{-2} \]

Calculating this gives:

\[ K_p = 0.5 \times 0.000315 = 0.0001575 \]

This result indicates the relationship between the pressure of the gases in equilibrium.

Example 2: Calculating Kp for a Decomposition Reaction

Consider the decomposition of dinitrogen tetroxide:

\[ 2N_2O_4(g) \rightleftharpoons 2NO_2(g) \]

Let’s say the equilibrium constant Kc is determined to be 4.0 at a specific temperature. To find Kp, we will again use the formula:

\[ K_p = K_c(RT)^{\Delta n} \]

For this reaction:

  • Moles of products = 2 (from 2NO2)
  • Moles of reactants = 2 (from 2N2O4)
  • \( \Delta n = 2 - 2 = 0 \)

At a temperature of 298 K, we have:

\[ K_p = 4.0 \times (0.0821 \times 298)^{0} = 4.0 \times 1 = 4.0 \]

In this case, Kp is equal to Kc, reflecting that there is no change in the number of moles of gas during the reaction.

Example 3: Calculating Kp Involving a Mixed Reaction

Let’s examine the reaction between carbon monoxide and hydrogen:

\[ CO(g) + H_2(g) \rightleftharpoons C_2H_6(g) \]

Suppose the equilibrium constant Kc is found to be 10.0 at a certain temperature. Here’s how to calculate Kp:

For this reaction:

  • Moles of products = 1 (from C2H6)
  • Moles of reactants = 2 (from CO and H2)
  • \( \Delta n = 1 - 2 = -1 \)

Using the formula:

\[ K_p = K_c(RT)^{\Delta n} \]

And assuming a temperature of 298 K:

\[ K_p = 10.0 \times (0.0821 \times 298)^{-1} \]

Calculating gives:

\[ K_p = 10.0 \times 0.0408 \approx 0.408 \]

This demonstrates how the equilibrium constants relate based on the change in volume, which is crucial for understanding gas-phase reactions.

These examples demonstrate the process of calculating Kp from Kc with varying complexities and contexts. Understanding these relationships is vital for anyone studying chemical equilibrium.