Question

What is \(K_{P}\) at \(1273^{\circ} \mathrm{C}\) for the reaction $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{CO}_{2}(g) $$ if \(K_{\mathrm{c}}\) is \(2.24 \times 10^{22}\) at the same temperature?

Short answer

\(K_P = 1.78 \times 10^{20}.\)

Step-by-step solution

Identify the Relationship Between Kc and Kp

To find \(K_{P}\) from \(K_{C}\), use the equation \(K_{P} = K_{C}(RT)^{\Delta n}\), where \(R\) is the gas constant (0.0821 L·atm/(mol·K)), \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in the number of moles of gas between products and reactants.

Convert Temperature to Kelvin

Convert the given temperature from Celsius to Kelvin using the formula \(T(K) = T(^{\circ}\mathrm{C}) + 273.15\). Thus, \(T = 1273 + 273.15 = 1546.15\, \text{K}.\) Approximate as \(T \approx 1546\, \text{K}.\)

Calculate Change in Moles (Δn)

Determine \(\Delta n\), which is the difference in moles of gaseous products and reactants. For the reaction \(2 \mathrm{CO}_{2}(g)\) minus \((2 \mathrm{CO}(g) + 1 \mathrm{O}_{2}(g))\), \(\Delta n = 2 - (2 + 1) = -1.\)

Insert Values into Kp Formula

Substitute \(K_{C} = 2.24 \times 10^{22}, \Delta n = -1, R = 0.0821\), and \(T = 1546\) K into the equation: \ \[ K_P = 2.24 \times 10^{22} \times (0.0821 \times 1546)^{-1} \]

Calculate Kp Result

Calculate \(K_P = 2.24 \times 10^{22} \times (0.0821 \times 1546)^{-1} \approx 2.24 \times 10^{22} \times 7.956 \times 10^{-3} \approx 1.78 \times 10^{20}.\)

Key concepts

Kc and Kp Relationship

Understanding the relationship between the equilibrium constants \(K_c\) and \(K_p\) is crucial when dealing with reactions involving gases. \(K_c\) is used for reactions in solution whereas \(K_p\) is applied to gaseous equilibria. The relationship between these two is expressed by the formula:\[ K_{P} = K_{C} (RT)^{\Delta n} \]where:

• \(R\) is the gas constant.
• \(T\) is the temperature in Kelvin.
• \(\Delta n\) is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants).

This relation highlights how changes in temperature or the number of moles can impact the equilibrium position depending on the states—gaseous or aqueous—of the reactants and products.

Conversion of Temperature to Kelvin

For chemical equilibria calculations involving gases, temperature must be expressed in Kelvin. The Kelvin scale is an absolute temperature scale starting at absolute zero, which is necessary for correct calculation of \(K_p\). Converting Celsius to Kelvin is straightforward:\[ T(K) = T(^{\circ}C) + 273.15 \]In our specific context, the conversion of \(1273^{\circ}C\) is done as follows:\[ T = 1273 + 273.15 = 1546.15\, \text{K} \]Rounding, for simplicity, gives \(T \approx 1546\,\text{K}\). This step is fundamental as equations involving the ideal gas constant \(R\) require temperatures in Kelvin.

Calculation of Change in Moles (Δn)

The change in moles \(\Delta n\) is critical when converting \(K_c\) to \(K_p\). It represents the difference in the number of moles of gaseous products and reactants and can be calculated from the balanced chemical equation. For the given reaction:\[ 2 \mathrm{CO}(g) + \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{CO}_2(g) \]we have:

• Moles of gaseous products: 2 (from \(\mathrm{CO}_2\))
• Moles of gaseous reactants: 2 + 1 = 3 (2 from \(\mathrm{CO}\) and 1 from \(\mathrm{O}_2\))

Thus, \(\Delta n = 2 - 3 = -1\). This negative value means the number of gas molecules decreases as the reaction proceeds from reactants to products.

Gas Constant R

The gas constant \(R\) is a key factor in the ideal gas law and related calculations for equilibria involving gases. Its value comes from the combination of basic gas properties:\[ R = 0.0821\, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) \]This specific version of \(R\) is used when pressure is in atmospheres and volume in liters. When you calculate \(K_p\) using \(K_c R T^{\Delta n}\), ensure you use the correct units for \(R\) to maintain consistency with other parameters, ensuring that \(T\) is in Kelvin and pressure units align with atmospheres. This constant helps tie together the thermal energy and pressure relations to the equilibrium state of the gases involved.