Question

Calculate \(K_{p}\) for each of the following equilibria: (a) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) ; K_{\mathrm{c}}=6.1 \times 10^{-3}\) at \(298 \mathrm{~K}\) (b) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) ; K_{\mathrm{c}}=2.4 \times 10^{-3}\) at \(1000 . \mathrm{K}\)

Short answer

For part (a), \( K_p \thickapprox 0.149 \). For part (b), \( K_p \thickapprox 3.56 \times 10^{-7} \).

Step-by-step solution

Understand the relationship between Kc and Kp

The relationship between the equilibrium constants Kc and Kp for a given reaction at a certain temperature is given by the equation: \[ K_p = K_c (RT)^{\bigtriangleup n} \] where \( R \) is the universal gas constant (0.0821 L atm / K mol), \( T \) is the temperature in Kelvin, and \( \bigtriangleup n \) is the change in the number of moles of gas (moles of gaseous products - moles of gaseous reactants).

Calculate \( \bigtriangleup n \) for Equilibrium (a)

For reaction (a): \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2 \text{NO}_2(g) \)The change in the number of moles of gas, \( \bigtriangleup n \) is: \[ \bigtriangleup n = (2) - (1) = 1 \]

Calculate \( K_p \) for Equilibrium (a)

We now use the equation derived in Step 1 to calculate \( K_p \) for reaction (a): \[ K_p = K_c (RT)^{\bigtriangleup n} \]Substituting the values: \( K_c = 6.1 \times 10^{-3} \ R = 0.0821 \ T = 298 \ \bigtriangleup n = 1 \) \[ K_p = 6.1 \times 10^{-3} (0.0821 \times 298)^{1} \] Perform the calculation: \[ K_p = 6.1 \times 10^{-3} \times 24.4558 \] \[ K_p \thickapprox 0.149 \]

Calculate \( \bigtriangleup n \) for Equilibrium (b)

For reaction (b): \( \text{N}_2(g) + 3 \text{H}_2(g) \rightleftharpoons 2 \text{NH}_3(g) \)The change in the number of moles of gas, \( \bigtriangleup n \) is: \[ \bigtriangleup n = (2) - (1 + 3) = 2 - 4 = -2 \]

Calculate \( K_p \) for Equilibrium (b)

We use the equation from Step 1 to calculate \( K_p \) for reaction (b): \[ K_p = K_c (RT)^{\bigtriangleup n} \]Substituting the values: \( K_c = 2.4 \times 10^{-3} \ R = 0.0821 \ T = 1000 \ \bigtriangleup n = -2 \) \[ K_p = 2.4 \times 10^{-3} (0.0821 \times 1000)^{-2} \] Perform the calculation: \[ K_p = 2.4 \times 10^{-3} \times (82.1)^{-2} \] \[ K_p = 2.4 \times 10^{-3} \times 0.0001484 \] \[ K_p \thickapprox 3.56 \times 10^{-7} \]

Key concepts

Kc and Kp relationship

The equilibrium constants, Kc and Kp, are crucial for understanding how gases behave in chemical reactions. Kc refers to the equilibrium constant in terms of concentration (molarity), while Kp is the equilibrium constant in terms of partial pressure. These two constants are related through the equation: \ K_p = K_c (RT)^{\Delta n} Here, \(R\) stands for the universal gas constant (0.0821 L atm / K mol), \(T\) is the temperature in Kelvin, and \(\Delta n\) represents the change in the number of moles of gas. This transformation is necessary because concentrations (Kc) and pressures (Kp) are directly connected through the ideal gas law (\(PV = nRT\)). By manipulating this relationship, you can convert between these constants to solve different types of equilibrium problems more flexibly and accurately.

Reaction Quotient

The reaction quotient, denoted as \(Q\), provides insight into the direction in which a reaction needs to proceed to reach equilibrium. While Kc and Kp are calculated using equilibrium concentrations or pressures, \(Q\) is determined using initial or non-equilibrium conditions. To calculate \(Q\), substitute the initial concentrations or pressures into the same formula used for equilibrium constants. Compare \(Q\) to \(K\) to determine the direction of the shift:

• If \(Q < K\), the reaction shifts to the right to produce more products.
• If \(Q > K\), the reaction shifts to the left to produce more reactants.
• If \(Q = K\), the reaction is at equilibrium.

Understanding \(Q\) helps predict and control how a reaction can be manipulated to reach or maintain equilibrium, making it a pivotal concept in chemical equilibria.

Universal Gas Constant

The universal gas constant, denoted as \(R\), forms the backbone of several key equations in chemistry, particularly in gas laws and equilibrium calculations. Its value is 0.0821 L atm / K mol. In the context of converting between Kc and Kp, \(R\) allows us to bridge the gap between concentrations and partial pressures. As seen in the equation \ K_p = K_c (RT)^{\Delta n} \(R\) helps ensure that the units are consistent, providing a smooth conversion process. This constant is also crucial in equations like the ideal gas law (\(PV = nRT\)) and others that describe gas behavior and reactions.

Delta n in Equilibrium Reactions

The term \(\Delta n\) signifies the change in the number of moles of gases in an equilibrium reaction. It is calculated by subtracting the moles of gaseous reactants from the moles of gaseous products. For example, in the reaction \ \text{N}_2 \text{O}_4(g) \rightleftrharpoons 2 \text{NO}_2(g) The \(\Delta n\) is calculated as \ \Delta n = (2) - (1) = 1. The value of \(\Delta n\) deeply influences how Kc is converted to Kp. When \(\Delta n\) is positive, the term \( (RT)^{\Delta n} \) increases the value of Kp relative to Kc, and vice versa. Keeping track of \(\Delta n\) is vital for correctly applying the relationship between Kc and Kp, guiding us to accurate predictions and understanding of chemical equilibrium behavior.