Question

At \(1100 \mathrm{K}, K_{\mathrm{p}}=0.25\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ What is the value of \(K\) at this temperature?

Short answer

The value of K at 1100 K for the given reaction is approximately 22.58.

Step-by-step solution

Calculate the change in moles of gas (∆n) in the balanced equation

First, we need to determine the change in the number of moles of gas in the balanced equation: Reactants: 2 moles of SO2(g) + 1 mole of O2(g) = 3 moles Products: 2 moles of SO3(g) ∆n = moles of products - moles of reactants ∆n = 2 - 3 = -1

Use the relation between Kp and Kc to find Kc

Now we'll use the relationship between Kp and Kc, which is: \( K_{p} = K_{c}(RT)^{\Delta n} \) Where R is the universal gas constant (0.0821 L atm/mol K), T is the temperature in Kelvin, and ∆n is the change in the number of moles of gas. We have Kp = 0.25, T = 1100 K, and ∆n = -1. Substituting these values into the formula, we get: \( 0.25 = K_{c}(0.0821\times1100)^{-1} \)

Determine the value of K (Kc)

To find Kc, we'll now solve the equation for Kc: \( K_{c} = \frac{0.25}{(0.0821\times1100)^{-1}} \) \( K_{c} = 0.25\times(0.0821\times1100) \) \( K_{c} = 0.25\times90.31 \) Calculating the value, we get: \( K_{c} = 22.5775 \) Therefore, the value of K at 1100 K is approximately 22.58.

Key concepts

Kp and Kc relationship

The relationship between the equilibrium constant for pressure, \(K_p\), and the equilibrium constant for concentration, \(K_c\), is important for understanding chemical equilibria, especially in gaseous reactions. In simple terms, \(K_p\) and \(K_c\) are related but not always equal. The difference arises because \(K_p\) depends on the pressures of gases, while \(K_c\) is concerned with their concentrations.

This relationship is governed by the formula:

• \( K_{p} = K_{c}(RT)^{\Delta n} \)

Here, \(\Delta n\) is the change in moles of gas, calculated as the moles of gaseous products minus the moles of gaseous reactants. \(R\) represents the gas constant, and \(T\) is the temperature in Kelvin.

To convert between these constants, knowing \(\Delta n\) is crucial because it reflects the shift in the number of gas particles, influencing the equilibrium pressure and concentration. This is especially significant for reactions where the number of moles of gas changes during the process.

Gas Constant

The gas constant, denoted as \(R\), plays a pivotal role in linking different properties of gases, such as pressure, volume, temperature, and the number of moles in equations like the ideal gas law. For equilibrium calculations, \(R\) is essential when relating \(K_p\) and \(K_c\).

The value of \(R\) used typically depends on the units being applied. In the context of equilibrium calculations and the conversion between \(K_p\) and \(K_c\), \(R\) is often given as:

• 0.0821 L atm/mol K

This value is specific for calculating relationships at particular conditions of temperature and pressure, such as the one given in our exercise at 1100 K. Using this gas constant allows us to accurately relate the pressure-based equilibrium constant, \(K_p\), with the concentration-based constant, \(K_c\), by accounting for the thermal and volumetric properties of gas molecules at equilibrium.

Equilibrium Constant Calculation

Calculating the equilibrium constant, whether \(K_p\) or \(K_c\), allows chemists to understand the position of equilibrium for a given chemical reaction. Here, we aim to find \(K_c\) when \(K_p\) is known, and we make use of the relation:

• \(K_{p} = K_{c}(RT)^{\Delta n}\)

Given that \(K_p = 0.25\), the temperature \(T = 1100\) K, and \(\Delta n = -1\), rearrange to solve for \(K_c\):

• First, calculate \((RT)^{\Delta n}\). Since \(\Delta n = -1\), compute \((0.0821 \times 1100)^{-1} = 1/90.31\).
• Substitute to find \(K_c\): \(K_c = 0.25 \times (90.31)\).
• The calculation gives \(K_c \approx 22.58\).

This process highlights how changes in the number of gas particles balance pressure and concentration influences, ultimately affecting the equilibrium state. In this exercise, the negative value of \(\Delta n\) reflects a decrease in gas molecules, which impacts the final calculation of \(K_c\).