Question

For a gaseous equilibrium: \(2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}), \mathrm{K}_{\mathrm{p}}\) has a value of \(1.8\) at \(700 \mathrm{~K} .\) What is the value of \(\mathrm{K}_{\mathrm{c}}\) for the equilibrium \(2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~A}\) at the same pressure: (a) \(0.031\) (b) \(1.3 \times 10^{-3}\) (c) \(44.4\) (d) 38

Short answer

The value of \( \mathrm{K}_{\mathrm{c}} \) is 0.031, which corresponds to option (a).

Step-by-step solution

Write the reaction equations

We start by writing the reverse reaction equation for which we need to find \(K_c\): \( 2 \mathrm{~B} + \mathrm{C} \rightleftharpoons 2 \mathrm{~A} \). This is the reverse of the given equilibrium equation.

Reverse reaction and equilibrium constant

The equilibrium constant \( K_p \) for the original reaction \( 2\mathrm{~A} \rightleftharpoons 2\mathrm{~B} + \mathrm{C} \) is given as 1.8. For the reverse reaction \( 2\mathrm{~B} + \mathrm{C} \rightleftharpoons 2\mathrm{~A} \), the equilibrium constant \( K_p' \) is the reciprocal of the original, so \( K_p' = \frac{1}{1.8} = 0.5556 \).

Relate Kc and Kp

The relation between \( K_c \) and \( K_p \) is \( K_p = K_c(RT)^{\Delta n} \), where \( R \) is the ideal gas constant (0.0821 L·atm/mol·K), \( T \) is the temperature in Kelvin (700 K), and \( \Delta n \) is the change in moles of gas when going from reactants to products.

Determine Δn for the reaction

For the reaction \( 2\mathrm{~B} + \mathrm{C} \rightleftharpoons 2\mathrm{~A} \), \( \Delta n = 2 - (2+1) = 2 - 3 = -1 \).

Calculate Kc for the reverse reaction

Now use the formula \[ K_c = \frac{K_p}{(RT)^{\Delta n}} \] to calculate \( K_c \) with \( K_p = 0.5556 \), \( R = 0.0821 \), \( T = 700 \), and \( \Delta n = -1 \). Thus, \( K_c = \frac{0.5556}{(0.0821 \times 700)^{-1}} \). Simplifying gives \( K_c = \frac{0.5556}{1/57.47} \approx 0.031 \).

Match your result with the given options

From the given options, \( K_c = 0.031 \) matches option (a).

Key concepts

Equilibrium Constants

In the study of chemical equilibrium, equilibrium constants play a crucial role in predicting the outcome of reversible reactions. For any reaction, we determine an equilibrium constant such as \( K_p \) or \( K_c \), depending on whether the system's concentration or pressure is being measured.
The equilibrium constant is a ratio of concentrations or partial pressures of products and reactants raised to the power of their coefficients.
Thus, for the reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K_c \) is defined as:

• \( K_c = \frac{{[C]^c[D]^d}}{{[A]^a[B]^b}} \)

However, when dealing with gases, we often use \( K_p \), which relies on partial pressures. The relationship between \( K_c \) and \( K_p \) is given by the equation:

• \( K_p = K_c(RT)^{\Delta n} \)

where \( \Delta n \) is the change in moles of gas and \( R \) is the ideal gas constant.

Le Chatelier's Principle

Le Chatelier's Principle helps us understand how a system at equilibrium responds to changes in concentration, pressure, and temperature. When a change is imposed on a system at equilibrium, this principle states that the system will adjust to counteract the imposed change and restore a new equilibrium.
For example,

• If the concentration of a reactant or product is changed, the system will shift towards the side that consumes the added substance, minimizing the change.
• Increasing pressure by reducing volume tends to favor the side of the reaction with fewer moles of gas.
• Temperature changes affect equilibria by shifting the balance depending on the reaction's endothermic or exothermic nature.

Thus, Le Chatelier’s Principle is an invaluable tool for predicting the direction in which an equilibrium will shift when disturbed.

Reaction Quotient

The Reaction Quotient \( Q \) is a snapshot of a reaction's position relative to equilibrium at any given moment. Unlike \( K \), which is calculated with equilibrium concentrations, \( Q \) compares the current state with the equilibrium state. For example,
in a reaction \( aA + bB \rightleftharpoons cC + dD \), the reaction quotient is:

• \( Q = \frac{{[C]^c[D]^d}}{{[A]^a[B]^b}} \)

To determine whether a reaction has reached equilibrium, we compare \( Q \) with \( K \):

• If \( Q = K \), the reaction is at equilibrium.
• If \( Q < K \), the reaction shifts towards the products to reach equilibrium.
• If \( Q > K \), the reaction shifts towards the reactants.

This concept is crucial for analyzing real-time progress and direction of chemical reactions.

Reversible Reactions

Reversible reactions are those that can proceed in both forward and backward directions. In a reversible reaction, the reactants convert to products and vice versa, as demonstrated by the double-headed arrow \( \rightleftharpoons \).
For a general reaction such as \( A + B \rightleftharpoons C + D \), equilibrium is achieved when the rate of the forward reaction equals the rate of the reverse reaction.
This dynamic situation means that even though the concentrations of reactants and products are constant at equilibrium, molecules are still reacting. Understanding reversible reactions is key to controlling chemical processes, maximizing yield, and minimizing waste.

• It allows chemists to manipulate conditions favorably to achieve desired outcomes in industrial settings, such as the Haber process for ammonia synthesis.

Hence, reversible reactions are fundamental to not only theoretical chemistry but practical applications as well.