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

Question

For the reaction \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}\) (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; it incorrectly represents the equilibrium constant expression.

Step 1: Understand the Reaction

The given reaction is \[\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(l)\] and it has an enthalpy change of \[\Delta_{r} \mathrm{H}=-170.8 \text{ kJ mol}^{-1}\].This negative enthalpy indicates that the reaction is exothermic, releasing heat when it proceeds to the right.

Step 2: Analyze Addition Effects

According to Le Chatelier's principle, adding more \(\mathrm{CH}_{4}(\mathrm{~g})\) or \(\mathrm{O}_{2}(\mathrm{~g})\) to the equilibrium mixture shifts the equilibrium to the right to consume these added reactants, forming more \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Thus, statement (a) is true.

Step 3: Identify Nature of the Reaction

The enthalpy change of the reaction, \(\Delta_{r} \mathrm{H}=-170.8 \text{ kJ mol}^{-1}\), confirms the reaction is exothermic, because a negative value indicates that heat is released during the reaction. Therefore, statement (b) is true.

Step 4: Consider Concentration Equivalence

For the reaction, since 1 mole of \(\mathrm{CO}_{2}\) and 2 moles of \(\mathrm{H}_{2} \mathrm{O}\) are produced, at equilibrium, their concentrations would not be equal under the same conditions. Therefore, statement (c) is true.

Step 5: Evaluate the Equilibrium Constant Expression

The equilibrium constant expression for gaseous reactions is derived from the balanced equation. The correct expression for the equilibrium constant \( K_p \) should take into account both reactants and products with gases appearing as pressure terms; for this reaction,The balanced reaction is:\[ K_p = \frac{ P_{\mathrm{CO}_{2}} }{ P_{\mathrm{CH}_{4}} \cdot (P_{\mathrm{O}_{2}})^2 } \]Statement (d) incorrectly represents \(K_p\) as it should include squared \(\mathrm{O}_{2}\) due to stoichiometry (i.e., \(\mathrm{O}_{2}\) is squared in the formula), so statement (d) is not true.

Key Concepts

Le Chatelier's Principle

Le Chatelier's Principle is a fundamental concept in chemical equilibrium. It helps us predict how a change in conditions can affect the equilibrium state of a chemical reaction. If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change. In simple terms, the system will try to "fix" the disturbance and restore balance.

Consider a reaction at equilibrium:

• Adding more reactants like \( ext{CH}_{4}( ext{g}) \) or \( ext{O}_{2}( ext{g}) \) will cause the system to shift towards the right. This means it will produce more products — \( ext{CO}_{2}( ext{g}) \) and \( ext{H}_{2} ext{O} ext{(l)} \).
• The principle applies under temperature, concentration, and pressure adjustments.

Keeping Le Chatelier's Principle in mind, it’s easier to understand how reactions can be manipulated by changing these factors.

Exothermic Reactions

An exothermic reaction is one that releases energy, typically in the form of heat, to its surroundings. This is indicated by a negative enthalpy change (\( ext{Δ}_{r} ext{H} \)). For the reaction, \( ext{CH}_{4}( ext{g}) + 2 ext{O}_{2}( ext{g}) \rightarrow \text{CO}_{2}( ext{g}) + 2 ext{H}_{2} ext{O(l)} \), the enthalpy is \( ext{Δ}_{r} ext{H} = -170.8 ext{ kJ mol}^{-1} \), signifying that it is indeed exothermic.

In the context of equilibrium, if you increase the temperature, this exothermic reaction will shift towards the left, favoring the reactants, as the system tries to absorb the added heat and minimize the disturbance. Conversely, lowering the temperature will shift the equilibrium to the right, favoring the formation of more products.

Equilibrium Constant Expression

The equilibrium constant expression (\( K_p \) for pressure, \( K_c \) for concentration) is a crucial element in understanding chemical equilibria. For a reaction involving gases:

\[\text{CH}_{4}( ext{g}) + 2 ext{O}_{2}( ext{g}) \rightleftharpoons \text{CO}_{2}( ext{g}) + 2 ext{H}_{2} ext{O(l)}\]
You write the expression for \( K_p \) as:
\[K_p = \frac{P_{ ext{CO}_{2}}}{P_{ ext{CH}_{4}} \times (P_{ ext{O}_{2}})^2} \]
This formula highlights an important stoichiometric concept: The coefficients in the balanced equation determine the powers to which each term is raised in the expression. Notice here, the reactant \( ext{O}_{2} \) is squared because two moles of it participate in the reaction.

Stoichiometry

Stoichiometry deals with the quantitative relationships in chemical reactions as dictated by the balanced equation. It's essential for determining how much of each substance is consumed or produced. In our reaction:

\[\text{CH}_{4}( ext{g}) + 2 ext{O}_{2}( ext{g}) \rightleftharpoons \text{CO}_{2}( ext{g}) + 2 ext{H}_{2} ext{O(l)}\]
We see that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water.

Using stoichiometry, we can predict the outcome of adding components or adjusting reaction conditions. For instance, if the amount of \( ext{CH}_{4} \) is increased, we can use stoichiometry to calculate the shift in equilibrium and determine how much more \( ext{CO}_{2} \) and \( ext{H}_{2} ext{O} \) will be produced to restore the equilibrium conditions. This highlights the direct link between stoichiometry and equilibrium dynamics.