Problem 34 - Chemistry: An Atoms-Focused Approach

Question

If the value of \(K_{c}\) for the following reaction is \(5 \times 10^{5}\) at \(298 \mathrm{K},\) what is the value of \(K_{\mathrm{p}}\) at \(298 \mathrm{K} ?\) $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g) $$

Step-by-Step Solution

Answer

Solution: The value of \(K_p\) for the given reaction at 298 K is approximately \(2.67 \times 10^{6}\).

Step 1: Write down the given data and equation

We are given the value of \(K_c = 5 \times 10^{5}\) at \(298~K\). The reaction is: $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{CO}_{2}(g) $$ Our goal is to find the value of \(K_p\).

Step 2: Find the change in moles of gas in the reaction \(\Delta n\)

To find \(\Delta n\), we can simply subtract the sum of the stoichiometric coefficients of the reactants from the sum of the stoichiometric coefficients of the products. $$ \Delta n = (2 - 2 - 1) = -1 $$

Step 3: Apply the relationship between \(K_p\) and \(K_c\)

Now that we have the value of \(\Delta n\), we can use the equation that relates \(K_p\) and \(K_c\): $$ K_p = K_c\left( \frac{RT}{P_0} \right)^{\Delta n} $$ Where \(R\) is the gas constant in appropriate units (\(R = \frac{0.0821~\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}}\)), \(T\) is the temperature (298 K), and \(P_0\) is the standard pressure (1 atm).

Step 4: Calculate the value of \(K_p\)

Plug in the given values, as well as the calculated \(\Delta n\), into the equation: $$ K_p = (5 \times 10^5)\left( \frac{(0.0821)(298)}{1} \right)^{-1} $$ Now, we can calculate \(K_p\): $$ K_p \approx 5 \times 10^{5} \times (0.0188)^{-1} = 2.67 \times 10^{6} $$ So the value of \(K_p\) at \(298~\mathrm{K}\) for the given reaction is approximately \(2.67 \times 10^{6}\).

Key Concepts

Reaction Stoichiometry

Understanding reaction stoichiometry is key to balancing chemical reactions and calculating various properties like equilibrium constants. Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. In the reaction provided, \[2 \mathrm{CO}(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{CO}_2(g)\], we see that:

• 2 moles of carbon monoxide (\(\mathrm{CO}\)) react with 1 mole of oxygen gas (\(\mathrm{O}_2\)).
• The result is 2 moles of carbon dioxide (\(\mathrm{CO}_2\)).

This means the ratio of \(\mathrm{CO}\) and \(\mathrm{O}_2\) in the reactants to \(\mathrm{CO}_2\) in the products is 2:1:2.
To compute changes in amount, look at the change in stoichiometric coefficients. By doing so, you can understand how the numbers of moles change as the reaction progresses. Here, we calculated \(\Delta n\) as \((2 - 2 - 1) = -1\), showing a decrease in one mole of gas on the reactant side compared to the product side.

Kp and Kc Relationship

The relationship between \(K_p\) and \(K_c\) is important when dealing with gaseous reactions. \(K_p\) is the equilibrium constant expressed in terms of partial pressures of gases, and \(K_c\) is in terms of concentrations. For the given reaction, we see that:

• \(K_c = 5 \times 10^{5}\) at 298 K.
• To convert \(K_c\) to \(K_p\), use the formula: \[K_p = K_c \left( \frac{RT}{P_0} \right)^{\Delta n}\]

This formula uses the universal gas constant \(R\), temperature \(T\), and change in number of moles \(\Delta n\).
The equation accounts for how pressure is related to concentration in a gaseous state, making it possible to switch between \(K_p\) and \(K_c\) as long as \(\Delta n\) is known. For our reaction, we find a \(K_p\) value of approximately \(2.67 \times 10^6\), indicating that as moles of gas decrease, this impacts the pressure, influencing \(K_p\) differently than \(K_c\).

Gas Constant

The gas constant \(R\) is a crucial factor in calculations involving gases. It's often used in the ideal gas law and related equations, describing the relationship between pressure, volume, temperature, and moles. For conversion between \(K_c\) and \(K_p\), it's essential to use the correct units for \(R\).

• In this exercise, \(R = 0.0821 \text{ L}\cdot\text{atm/mol}\cdot\text{K}\).
• This is the typical unit used for such conversions when pressure is in atmospheres and volume in liters.

The role of \(R\) helps standardize our calculations, ensuring that temperature and pressure conditions are appropriately addressed.
Temperature, represented as \(T\), should be converted into Kelvin to maintain consistency with \(R\)'s units.
This means, at 298 K, our use of \(R\) allows for accurate predictions of \(K_p\) based on \(K_c\), maintaining the applicability of thermodynamic principles across varying reaction conditions.