Problem 51 - A Complete Resource Book in Chemistry for JEE Main

Question

For the reaction $$ \mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$ \(\Delta_{r} \mathrm{H}=-170.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) Which of the following statements is not true? (a) addition of \(\mathrm{CH}_{4}(\mathrm{~g})\) or \(\mathrm{O}_{2}(\mathrm{~g})\) at equilibrium will cause a shift to the right (b) the reaction is exothermic (c) at equilibrium, the concentrations of \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}\) (1) are not equal (d) the equilibrium constant for the reaction is given by \(\mathrm{K}_{\mathrm{p}}=\frac{\left[\mathrm{CO}_{2}\right]}{\left[\mathrm{CH}_{4}\right]\left[\mathrm{O}_{2}\right]}\)

Step-by-Step Solution

Answer

Statement (d) is not true.

Step 1: Identify the Type of Reaction

Analyze the given reaction equation: \[ \text{CH}_{4}(\text{g}) + 2 \text{O}_{2}(\text{g}) \rightleftharpoons \text{CO}_{2}(\text{g}) + 2 \text{H}_{2} \text{O}(\text{l}) \]The reaction is a combustion reaction of methane in the presence of oxygen, producing carbon dioxide and water.

Step 2: Recognize the Reaction Enthalpy

The given enthalpy change \( \Delta_{r} \text{H} = -170.8 \text{ kJ mol}^{-1} \) is negative, indicating that the reaction is exothermic. Thus, statement (b) is true.

Step 3: Assess Le Chatelier's Principle

According to Le Chatelier's Principle, adding \( \text{CH}_{4}(\text{g}) \) or \( \text{O}_{2}(\text{g}) \) will disturb the equilibrium and cause the system to shift to the right to consume the added reactants. Therefore, statement (a) is true.

Step 4: Evaluate Concentration Equality

At equilibrium, the concentrations of products and reactants are constant, but not necessarily equal. Since \( \text{CO}_{2} \) is a gas and \( \text{H}_{2} \text{O} \) is a liquid in this reaction, their concentrations are calculated differently. Thus, statement (c) is likely true, but must be considered carefully.

Step 5: Analyze the Equilibrium Constant Expression

For the reaction, the equilibrium expression in terms of partial pressures, \( K_p \), does not include \( \text{H}_{2} \text{O} \) (l), as its activity is considered to be 1. Thus, \( K_p \) should be:\[ K_p = \frac{[\text{CO}_2]}{[\text{CH}_4][\text{O}_2]^2} \]Statement (d) appears incorrect because it misses the square of \( [\text{O}_2] \), crucial for balancing the stoichiometry.

Key Concepts

Le Chatelier's Principle

In chemical reactions like the combustion of methane, "Le Chatelier's Principle" helps us to predict how the system will behave when subjected to changes. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, such as concentration, pressure, or temperature, the position of equilibrium shifts to counteract the change and restore balance.

For instance, if we increase the concentration of reactants such as \(\text{CH}_4(\text{g})\) or \(\text{O}_2(\text{g})\), the system will shift to the right. This means that more products, \(\text{CO}_2(\text{g})\) and \(\text{H}_2\text{O}(\text{l})\), will be formed to decrease the added reactants. Similarly, a decrease in reactants will push the equilibrium to the left, favoring the formation of reactants again.

This principle also applies when temperature is changed. Since this reaction is exothermic, a rise in temperature will push the equilibrium position to the left, favoring the endothermic backward reaction, as the system tries to absorb the increased heat. Conversely, lowering the temperature will favor the forward exothermic reaction. This knowledge helps in predicting how different conditions affect chemical equilibria.

Reaction Enthalpy

The concept of 'Reaction Enthalpy' refers to the heat change that occurs during a chemical reaction. In our given reaction, the formation of \(\text{CO}_2(\text{g})\) and \(\text{H}_2\text{O}(\text{l})\) from \(\text{CH}_4(\text{g})\) and \(\text{O}_2(\text{g})\) has an enthalpy change, \(\Delta_{r} \text{H} = -170.8 \text{ kJ mol}^{-1}\).

This negative sign indicates that the reaction releases heat, classifying it as exothermic. Exothermic reactions are characterized by the release of energy to the surroundings, usually making the reaction products more stable than the reactants.

Understanding reaction enthalpy is critical, as it influences the reaction conditions. For instance, if heat is removed from an exothermic reaction, it can shift the equilibrium position to favor the formation of products, as predicted by Le Chatelier’s Principle. Reaction enthalpy also affects how we design industrial processes to either harness the released energy or control the reaction conditions effectively.

Equilibrium Constant

Understanding the 'Equilibrium Constant', particularly \(K_p\) in gaseous reactions, is crucial in evaluating the state of the reaction at equilibrium. It reflects the ratio of the partial pressures of the products to that of reactants, adjusted for their stoichiometric coefficients.

For the given combustion reaction of methane, where water is in a liquid state that is not included in the \(K_p\) calculations due to its constant activity, the \(K_p\) equation is formed by:

• Including only gaseous components – \(\text{CO}_2(\text{g})\) and \(\text{CH}_4(\text{g})\).
• The correct stoichiometric balancing requiring the square of \([\text{O}_2(\text{g})]\) in the equation because two moles of \(\text{O}_2\) are consumed per mole of \(\text{CH}_4\).

The accurate expression thus becomes:
\[ K_p = \frac{[\text{CO}_2]}{[\text{CH}_4][\text{O}_2]^2} \]

Comparing actual \(K_p\) values with calculated ones helps us understand the extent to which a reaction favors products or reactants, guiding us in optimizing conditions for desired outputs in industrial applications.