Question

For a gaseous reaction \(2 \mathrm{~A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\), the par- tial pressures of \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) at equilibrium are \(0.5\), \(0.8,0.7\) and \(1.2 \mathrm{~atm}\). The value of \(\mathrm{K}_{\mathrm{p}}\) for this reaction is: (a) \(2.4 \mathrm{~atm}\) (b) \(6.2 \mathrm{arm}^{-2}\) (c) \(4.2 \mathrm{~atm}^{-1}\) (d) \(8.4 \mathrm{~atm}^{-3}\)

Short answer

The value of \(K_p\) is \(4.2 \mathrm{~atm}^{-1}\), corresponding to option (c).

Step-by-step solution

Write the Expression for Kp

For the reaction \(2 \text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}\), the equilibrium constant \(K_p\) is given by the expression \[ K_p = \frac{P_C \cdot P_D}{P_A^2 \cdot P_B} \] where \(P_A, P_B, P_C,\) and \(P_D\) represent the partial pressures of A, B, C, and D, respectively.

Substitute Given Values

Substitute the given partial pressures into the \(K_p\) expression: \(P_A = 0.5\) atm, \(P_B = 0.8\) atm, \(P_C = 0.7\) atm, \(P_D = 1.2\) atm. The equation becomes \[ K_p = \frac{0.7 \times 1.2}{(0.5)^2 \times 0.8} \]

Calculate the Numerator

Calculate the product \( 0.7 \times 1.2 = 0.84 \). This is the numerator of our \(K_p\) expression.

Calculate the Denominator

Calculate \((0.5)^2 \times 0.8\). First square 0.5 to get 0.25, then multiply by 0.8 to obtain \(0.25 \times 0.8 = 0.2\). This is the denominator of our \(K_p\) expression.

Divide Numerator by Denominator

Divide the numerator by the denominator to find \(K_p\): \[ K_p = \frac{0.84}{0.2} = 4.2 \text{ atm}^{-1} \]

Choose the Correct Option

Compare the calculated \(K_p\) value with the given options: \(4.2 \text{ atm}^{-1}\). This matches option (c).

Key concepts

Partial Pressure

In the context of gaseous reactions, partial pressure is a crucial concept. Each gas in a mixture contributes to the total pressure in proportion to its amount according to the ideal gas law. When dealing with reactions at equilibrium, the partial pressure of each gas plays a significant role in calculating the equilibrium constant, known as the equilibrium constant in terms of pressure, or \(K_p\). In the given reaction \(2 \text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}\), the partial pressures provided were \(P_A = 0.5\, \text{atm}\), \(P_B = 0.8\, \text{atm}\), \(P_C = 0.7\, \text{atm}\), and \(P_D = 1.2\, \text{atm}\). These values are vital inputs for the calculation of \(K_p\).

To understand the relationship more clearly, visualize each partial pressure as a fraction of the total pressure, which helps to measure how much each gas contributes individually to the system's total pressure. Remember that in an equilibrium system, these pressures remain constant, assuming the temperature remains unchanged.

Gaseous Reactions

Gaseous reactions involve reactants and products in the gas phase. The simplicity of measuring gases by their pressure makes these reactions easier to quantify using principles like ideal gas law. In reaction \(2 \text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}\), represented in a balanced chemical equation, it indicates the conversion of reactants \(A\) and \(B\) to products \(C\) and \(D\) and vice versa.

Understanding gaseous reactions requires considering moles of gas and their changes during reaction. For each substance in our example reaction:

• Every 2 moles of \(A\) react with 1 mole of \(B\) to form 1 mole of \(C\) and 1 mole of \(D\).
• Changes in moles result in changes in pressure, as described by the ideal gas law \(PV = nRT\).

Utilizing partial pressures to represent concentration provides a more practical approach than moles for gases at constant temperature and volume.

Chemical Equilibrium

Chemical equilibrium in a gas phase reaction occurs when the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations, or partial pressures, of reactants and products remain unchanged over time. Understanding chemical equilibrium allows us to predict the conditions under which a reaction will proceed and to what extent.

For the equilibrium described by \(2 \text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}\), it means that under equilibrium conditions:

• The rates at which \(A\) and \(B\) become \(C\) and \(D\) are perfectly balanced by the rates at which the reverse is happening.
• The system is in dynamic equilibrium; the reactions occur continuously, but there is no net change in the composition of the system.

While the pressures given determine the equilibrium state, changes in temperature, pressure or concentration can shift the equilibrium, as defined by Le Chatelier's Principle.

Reaction Quotient

The reaction quotient \(Q_p\) is a valuable tool for predicting the direction in which a reaction must proceed to reach equilibrium when not at equilibrium. It is constructed using the same formula as the equilibrium constant \(K_p\), differing only in that its values can be taken at any given point and not necessarily at equilibrium.

For our reaction \(2 \text{A} + \text{B} \rightleftharpoons \text{C} + \text{D}\), the reaction quotient is given as:

• \[ Q_p = \frac{P_C \cdot P_D}{P_A^2 \cdot P_B} \]
• By comparing \(Q_p\) to \(K_p\), we can tell whether the reaction needs to proceed in the forward or reverse direction to achieve equilibrium.

If \(Q_p < K_p\), the reaction moves forward, increasing product concentration. If \(Q_p > K_p\), the reaction shifts backward, increasing reactant concentration. Understanding \(Q_p\) helps balance chemical systems effectively.