Problem 36 - Chemistry: An Atoms-Focused Approach

Question

For which of the following reactions are the values of \(K_{c}\) and \(K_{\mathrm{p}}\) different? a. \(\operatorname{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\) b. \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) c. \(2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)\)

Step-by-Step Solution

Answer

Answer: For all the reactions - (a), (b), and (c) - the values of \(K_c\) and \(K_p\) are different.

Step 1: Relationship between Equilibrium Constants \(K_c\) and \(K_p\)

The relationship between \(K_c\) and \(K_p\) for a reaction is given by the formula: $$K_p = K_c(RT)^{\Delta n}$$ Where \(K_p\) is the equilibrium constant in terms of partial pressures, \(K_c\) is the equilibrium constant in terms of concentrations, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in the number of moles of gas during the reaction (moles of products - moles of reactants).

Step 2: Calculate \(\Delta n\) for Reaction a

For reaction a: $$\operatorname{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ The moles of gaseous reactants are 1, while the moles of gaseous products are 2: $$\Delta n = 2 - 1 = 1$$

Step 3: Calculate \(\Delta n\) for Reaction b

For reaction b: $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ The moles of gaseous reactants are 3 while the moles of gaseous products are 2: $$\Delta n = 2 - 3 = -1$$

Step 4: Calculate \(\Delta n\) for Reaction c

For reaction c: $$2 \mathrm{O}_{3}(g) \rightleftharpoons 3 \mathrm{O}_{2}(g)$$ The moles of gaseous reactants are 2 while the moles of gaseous products are 3: $$\Delta n = 3 - 2 = 1$$

Step 5: Determine if \(K_c\) and \(K_p\) are different

In each case, the \(\Delta n\) values that we calculated during each reaction were 1, -1, and 1, respectively. We can substitute these values into the relationship between \(K_c\) and \(K_p\): For reaction a: $$K_p = K_c(RT)^1$$ For reaction b: $$K_p = K_c(RT)^{-1}$$ For reaction c: $$K_p = K_c(RT)^1$$ If \(\Delta n = 0\), \(K_p\) and \(K_c\) would be the same; however, all reactions have a non-zero \(\Delta n\). Therefore, for all the reactions - (a), (b), and (c) - the values of \(K_c\) and \(K_p\) are different.

Key Concepts

Kc vs Kp

In the study of chemical equilibria, we encounter two types of equilibrium constants: \(K_c\) and \(K_p\). Both serve as metrics for the equilibrium state, yet they apply to different scenarios. \(K_c\) is used when concentrations are expressed in molarity (moles per liter), thus more suitable for solutions. On the other hand, \(K_p\) is utilized when dealing with gases, as it relates to the partial pressures of gases. This is because pressure is an effective measure of concentration for gaseous substances.

The relationship between \(K_c\) and \(K_p\) is illustrated by the equation:

• \(K_p = K_c (RT)^{\Delta n}\)

Here, \(R\) represents the ideal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) indicates the change in moles of gas. This equation highlights that the difference between \(K_c\) and \(K_p\) depends on both temperature and changes in mole numbers. If \(\Delta n\) is zero, \(K_c\) equals \(K_p\); otherwise, they differ. This difference must be considered when analyzing reactions involving gaseous components.

Partial Pressure

Partial pressure refers to the pressure exerted by a particular gas in a mixture of gases. It's an important concept when calculating \(K_p\) since \(K_p\) involves the pressures of each participating gas. According to Dalton's Law, the total pressure exerted by a gas mixture equals the sum of the partial pressures of individual gases present.

We can represent partial pressure as:

• \(P_i = \chi_i \cdot P_{\text{total}}\)

where \(P_i\) is the partial pressure of a specific gas, \(\chi_i\) is its mole fraction, and \(P_{\text{total}}\) denotes the total pressure of the gas mixture.

Understanding partial pressures is crucial for calculating \(K_p\) because it involves raising each partial pressure to the power of its stoichiometric coefficient in the balanced chemical equation. This method allows chemists to determine how gases participate in and reach equilibrium under various pressures.

Change in Moles of Gas

The change in moles of gas, denoted as \(\Delta n\), is a fundamental aspect of understanding the differences between \(K_c\) and \(K_p\). It represents the net change in the number of moles of gaseous reactants and products in a chemical reaction.

• \(\Delta n = \text{moles of products} - \text{moles of reactants}\)

This change directly influences the relationship between \(K_c\) and \(K_p\). For example, if a reaction produces more moles of gas than it consumes (like in Reaction c), \(\Delta n\) is positive, which affects \(K_p\)'s value relative to \(K_c\) as expressed by the formula \((RT)^{\Delta n}\).

Each reaction step requires calculating \(\Delta n\) to understand how it impacts the corresponding equilibrium constants. In scenarios where \(\Delta n\) is zero, \(K_c\) and \(K_p\) are equal, maintaining simplicity in calculations. Recognizing these changes helps us predict and explain variations in reactions involving gases, particularly in shifting equilibrium at different pressures.