Problem 22 - Chemistry The Central Science In Si Units

Question

Calculate \(K_{c}\) at \(900 \mathrm{~K}\) for \(2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\) if \(K_{p}=0.0572\) at this temperature.

Step-by-Step Solution

Answer

The equilibrium constant \(K_c\) at 900 K is 4.224.

Step 1: Understanding the Relationship

First, recognize that you want to relate the equilibrium constant in terms of pressure, \(K_p\), to that in terms of concentration, \(K_c\). This relationship is given by the equation \(K_p = K_c(RT)^{\Delta n}\), where \(\Delta n\) is the change in moles of gas, \(R\) is the universal gas constant in L·atm/mol·K, and \(T\) is the temperature in Kelvin.

Step 2: Determine Δn

Find \(\Delta n\), which is the difference in the number of moles of gaseous products and reactants. In the reaction \(2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_2(g) + \mathrm{C}(s)\), the change in moles for gases is calculated as follows:\[ \Delta n = (1 - 2) = -1 \]Since carbon \( \text{C} \) is a solid, it does not contribute to \( \Delta n \).

Step 3: Substitute Known Values

Using the relationship \(K_p = K_c(RT)^{\Delta n}\), substitute the known values:- \(K_p = 0.0572\)- \(R = 0.0821 \text{ L·atm/mol·K}\)- \(T = 900 \text{ K}\)- \(\Delta n = -1\)We need to solve for \(K_c\).

Step 4: Calculate K_c

Rearrange to find \(K_c\):\[ K_c = \frac{K_p}{(RT)^{\Delta n}} \]Substitute the values:\[ K_c = \frac{0.0572}{(0.0821 \times 900)^{-1}} \]Calculate \(RT\):\[ RT = 0.0821 \times 900 = 73.89 \]Now plug this back into our equation:\[ K_c = \frac{0.0572}{(73.89)^{-1}} \]And finally, compute \(K_c\):\[ K_c = 0.0572 \times 73.89 = 4.224 \]

Step 5: Conclusion

Therefore, the equilibrium constant \(K_c\) at 900 K for the reaction is 4.224.

Key Concepts

Chemical Equilibrium

In a chemical reaction, equilibrium refers to the state where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time. Equilibrium doesn't mean the amounts are equal, but rather that their rates of change are balanced.

• Reactions at equilibrium are dynamic, meaning reactions continue to occur, but there's no net change in concentration.
• Equilibrium is affected by changes in conditions, such as temperature, pressure, or concentration of reactants/products.
• The equilibrium constant, denoted as either \( K_c \) or \( K_p \), quantifies the position of the reaction's balance. A larger value of \( K \) indicates a product-favored equilibrium, whereas a smaller value points to a reactant-favored state.

Understanding chemical equilibrium helps us predict how changes in conditions can affect a reaction, guiding us in controlling chemical processes effectively.

Gaseous Reactions

Gaseous reactions involve reactions where one or more reactants or products are in the gas phase. These reactions can be influenced by changes in temperature, pressure, and volume due to the gases' ability to expand and compress.

• Gases are often involved in equilibria because their concentrations (or pressures) change with shifts in conditions.
• The behavior of gases in these reactions can be described using the ideal gas law equation: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is temperature in Kelvin.
• Soft solids, like carbon in this exercise, don't impact the gaseous equilibrium constant, as they don't contribute to the change in mole numbers.

Gaseous reactions are crucial in many industrial processes, and understanding their equilibrium behavior is essential for designing and optimizing reactors.

Relationship between Kc and Kp

The relationship between \( K_c \) and \( K_p \) is integral when dealing with reactions involving gases, as these constants express equilibrium in different terms:
\[ K_p = K_c(RT)^{\Delta n} \] Here, \( \Delta n \) represents the difference in moles of gaseous products and reactants. This relation allows us to convert between the concentration-based \( K_c \) and the pressure-based \( K_p \), depending on what's most convenient or required for analysis.

• \( R \) is the gas constant (0.0821 L·atm/mol·K), which ensures the units align when converting \( K_c \) to \( K_p \).
• \( \Delta n \) is critical as it accounts for the difference in moles of gases, influencing the calculation of \( K_p \) from \( K_c \).
• If \( \Delta n = 0 \), then \( K_c \) and \( K_p \) are equal as the effect of temperature cancels out.

This relationship is essential for solving problems involving changes in equilibrium constants at varying conditions.