Problem 200 - A Complete Resource Book in Chemistry for JEE Main

Question

For the following three reactions \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\), equilibrium constants are given: [2008] (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{1}\)(b) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{2}\) (c) \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2}(\mathrm{~g}) ; \mathrm{K}_{3}\) Which of the following relation is correct? (a) \(\mathrm{K}_{1} \sqrt{\mathrm{K}}_{2}=\mathrm{K}_{2}\) (b) \(\mathrm{K}_{2} \mathrm{~K}_{3}=\mathrm{K}_{1}\) (c) \(\mathrm{K}_{3}=\mathrm{K}_{1} \mathrm{~K}_{2}\) (d) \(\mathrm{K}_{3} \cdot \mathrm{K}_{2}^{3}=\mathrm{K}_{1}^{2}\)

Step-by-Step Solution

Answer

The correct relation is (c) \( \mathrm{K}_{3} = \mathrm{K}_{1} \mathrm{~K}_{2} \).

Step 1: Write the Equilibrium Constant for Each Reaction

We will write out the equilibrium expression for each reaction based on the given equations:- For Reaction (a), the equilibrium constant is \( \mathrm{K}_{1} = \frac{[\mathrm{CO}_2][\mathrm{H}_2]}{[\mathrm{CO}][\mathrm{H}_2 \mathrm{O}]} \).- For Reaction (b), the equilibrium constant is \( \mathrm{K}_{2} = \frac{[\mathrm{CO}][\mathrm{H}_2]^3}{[\mathrm{CH}_4][\mathrm{H}_2 \mathrm{O}]} \).- For Reaction (c), the equilibrium constant is \( \mathrm{K}_{3} = \frac{[\mathrm{CO}_2][\mathrm{H}_2]^4}{[\mathrm{CH}_4][\mathrm{H}_2 \mathrm{O}]^2} \).

Step 2: Derive the Relationship Between Reactions

To find a relationship, we arrange the reactions such that they can be summed to form Reaction (c). Note that Reaction (c) is obtained by combining Reactions (a) and (b).- If we consider adding Reaction (a) and Reaction (b), we need to add these equations such that the result is Reaction (c): - \( [\mathrm{CH}_4] + 2[\mathrm{H}_2 \mathrm{O}] \rightleftharpoons [\mathrm{CO}_2] + 4[\mathrm{H}_2]\).- From this, the equilibrium constant \( \mathrm{K}_{3} \) can be expressed as the product of the equilibrium constants from Reactions (a) and (b): \( \mathrm{K}_{3} = \mathrm{K}_{1} \times \mathrm{K}_{2} \).

Step 3: Verify Potential Relationships

Consider option (c) \( \mathrm{K}_{3} = \mathrm{K}_{1} \mathrm{~K}_{2} \), given our previous derivation, this is correct as it represents the combination of the reactions to form Reaction (c).- Check calculations by substituting the equilibrium constant expressions derived: - \( \mathrm{K}_{3} = \left( \frac{[\mathrm{CO}_2][\mathrm{H}_2]}{[\mathrm{CO}][\mathrm{H}_2 \mathrm{O}]} \right) \times \left( \frac{[\mathrm{CO}][\mathrm{H}_2]^3}{[\mathrm{CH}_4][\mathrm{H}_2 \mathrm{O}]} \right) = \frac{[\mathrm{CO}_2][\mathrm{H}_2]^4}{[\mathrm{CH}_4][\mathrm{H}_2 \mathrm{O}]^2} \).-The expression confirms the relation \( \mathrm{K}_{3} = \mathrm{K}_{1} \times \mathrm{K}_{2} \).

Key Concepts

Equilibrium Constants

When discussing chemical equilibrium, one central concept to understand is the equilibrium constant, often represented as \(K\). This constant is crucial because it helps us quantify the balance between reactants and products in a reversible chemical reaction at equilibrium. In simple terms, it tells us how far a reaction will proceed before reaching equilibrium.
The value of an equilibrium constant is derived from the concentrations of the products and reactants, presented in a specific ratio. For example, in a reaction where substance \(A\) reacts with substance \(B\) to form \(C\) and \(D\), the equilibrium constant expression would be \(K = \frac{[C][D]}{[A][B]}\). It's important to note that each concentration is raised to the power of its corresponding stoichiometric coefficient in the balanced equation.
Key points about equilibrium constants include:

• An equilibrium constant is specific to a particular reaction and depends on temperature.
• A large \(K\) indicates that the formation of products is favored; whereas a small \(K\) suggests that the reactants are favored.
• By manipulating equilibrium constants, we can understand the relationship and balance between multiple reactions, as seen in how \(K_3\) from the exercise is a product of \(K_1\) and \(K_2\).

Having a solid grasp on equilibrium constants allows us to predict the direction and extent of chemical reactions which is crucial for many practical applications in chemistry and industry.

Reaction Stoichiometry

Stoichiometry in reactions is all about the balance and proportion of different substances that participate in chemical transformations. In our exercise, the stoichiometry of the reactions tells us the amounts of reactants needed and products formed, which is critical for understanding chemical equilibrium.
The stoichiometric coefficients in a balanced chemical equation relate to the number of moles of each substance involved. For instance, in the reaction \(\mathrm{CH}_4 + 2\mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{CO}_2 + 4\mathrm{H}_2\), the coefficients \(1, 2, 1,\) and \(4\) tell us the proportion of molecules or moles required and produced during the reaction.
Key stoichiometry points in equilibrium reactions include:

• Each coefficient is crucial for constructing an accurate equilibrium constant expression.
• The stoichiometry helps determine how products are formed relative to the original reactants, affecting the equilibrium condition.
• In complex reactions, such as in the given problem, understanding the stoichiometry helps to identify how different reactions can combine or transform into one another.

Overall, understanding stoichiometry is indispensable for predicting how much reactant is consumed and product is produced, paving the way for understanding the efficiency and outcome of chemical reactions.

Thermodynamics in Chemistry

Thermodynamics is the branch of chemistry that deals with the energy changes involved in chemical processes. It helps explain why reactions happen, how they occur, and what equilibrium state they tend towards.
In the context of chemical equilibrium, thermodynamics provides insight into how energy influences the position of equilibrium. Through the Gibbs free energy change (\(\Delta G\)), it provides a criterion for spontaneity of reactions:

• If \(\Delta G\) is negative, a reaction will proceed spontaneously in the forward direction.
• If \(\Delta G\) is positive, the reaction is spontaneous in the reverse direction.
• A \(\Delta G\) of zero indicates that the system is at equilibrium.

Thermodynamics also plays a role in determining the equilibrium constant through the relation \(\Delta G = -RT \ln K\), where \(R\) is the gas constant, \(T\) is temperature in Kelvin, and \(\ln\) is the natural logarithm. This equation tells us that equilibrium constants are inherently tied to the energy changes in a reaction, explaining why they vary with temperature.
By mastering the thermodynamic principles underpinning chemical reactions, one gains a deeper understanding of how reactions progress and reach equilibrium states, allowing for more effective predictions and control of chemical processes.