Problem 75 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction \(\mathrm{C}_{(s s}+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})\), the par- tial pressures of \(\mathrm{CO}\), and \(\mathrm{CO}\) are 4 and 8 atm respectively. The value of \(K_{ }^{\prime}\) for this reaction is (a) \(14 \mathrm{~atm}\) (b) \(16 \mathrm{~atm}\) (c) \(18 \mathrm{~atm}\) (d) \(12 \mathrm{~atm}\)

Step-by-Step Solution

Answer

The calculated equilibrium constant \(K_p\) is 2 atm, not matching any given choices.

Step 1: Write the Equilibrium Expression

For the given reaction \(C_{(s)} + CO_{2(g)} \rightleftharpoons 2 CO_{(g)}\), we write the equilibrium constant expression \(K_p\) as: \[ K_p = \frac{{(P_{CO})^2}}{{P_{CO_2}}} \] where \(P_{CO}\) is the partial pressure of carbon monoxide and \(P_{CO_2}\) is the partial pressure of carbon dioxide.

Step 2: Plug in Given Values

From the problem, \(P_{CO} = 4 \text{ atm}\) and \(P_{CO_2} = 8 \text{ atm}\). Substitute these values into the equilibrium expression:\[ K_p = \frac{{4^2}}{{8}} \]

Step 3: Solve the Mathematical Expression

Calculate the numerical value of \(K_p\):\[ K_p = \frac{{16}}{{8}} = 2 \text{ atm} \]

Step 4: Compare with Given Choices

The possible choices for the equilibrium constant \(K_p\) given are:(a) \(14 \text{ atm}\)(b) \(16 \text{ atm}\)(c) \(18 \text{ atm}\)(d) \(12 \text{ atm}\)None of these options match the calculated \(K_p\) of 2 atm. More information or an adjustment may be needed.

Key Concepts

Equilibrium Constant Expression

In chemical equilibrium, the equilibrium constant expression is crucial as it quantifies the position of equilibrium. For a reaction like \(C_{(s)} + CO_{2(g)} \rightleftharpoons 2 CO_{(g)}\), this expression helps us understand how concentrations or partial pressures relate at equilibrium. The equilibrium constant, denoted as \(K_p\) when we talk about partial pressures, expresses this relationship. For gaseous reactions, it simplifies to a fraction where the numerator is the product of the partial pressures of the products raised to the power of their stoichiometric coefficients. The denominator features the partial pressures of the reactants, also raised to their stoichiometric coefficients.
In our example, the equilibrium constant expression is \(K_p = \frac{{(P_{CO})^2}}{{P_{CO_2}}}\). It's important to note here that pure solids like \(C_{(s)}\) do not appear in the equilibrium expression because their activity is considered as 1. Calculating \(K_p\) involves plugging in the known partial pressures and solving the mathematical expression, which gives us insights into the reaction's dynamics.

Partial Pressure

Partial pressure is a concept vital in understanding reactions involving gases. It refers to the pressure exerted by a single type of gas in a mixture of gases. For a reaction like \(C_{(s)} + CO_{2(g)} \rightleftharpoons 2 CO_{(g)}\), knowing the partial pressures of \(CO_2\) and \(CO\) is necessary to use the equilibrium constant expression.
Partial pressures can be thought of as individual contributions to the total pressure, allowing us to predict which way a reaction is likely to shift. If we add a gas to the system, increasing its partial pressure, the system will try to relieve this stress by shifting the reaction to favor the consumption of the added gas, according to Le Chatelier's Principle.
In practical terms, partial pressures are typically measured in atmospheres (atm) and can often be controlled in laboratory settings, making them an indispensable tool in manipulating and studying chemical equilibria.

Reversible Reactions

Reversible reactions are those that can proceed in both forward and backward directions simultaneously. A state of equilibrium is achieved when the rate of the forward reaction equals that of the backward reaction, maintaining the concentrations of reactants and products constant over time. The reaction \(C_{(s)} + CO_{2(g)} \rightleftharpoons 2 CO_{(g)}\) is an example of a reversible reaction.
In reversible reactions, it's important to recognize the dynamic nature of equilibrium: reactions aren't stopped but rather balanced. Changes in conditions such as temperature, pressure, or concentrations can disturb the equilibrium, prompting the reaction to shift in the direction that mitigates the change, an explanation provided by Le Chatelier's Principle.
Reversible reactions are central to many natural and industrial processes. Understanding their nature and how to manipulate their equilibria is crucial for fields ranging from synthetic chemistry to environmental science, and even in processes like the Haber process for ammonia synthesis.