N2O4(g) ↔ 2NO2(g) ΔH = +ve In the reaction above, an increase in temperature will

Analysis of the Question

This question tests your understanding of Le Chatelier’s Principle and how temperature changes affect the equilibrium constant ($K$) of a reversible reaction.

Concept Explanation

Le Chatelier’s Principle states that if a dynamic equilibrium is disturbed by changing the conditions (like temperature, pressure, or concentration), the position of equilibrium shifts to counteract the change.

• Endothermic Reactions ($\Delta H = +ve$): These reactions absorb heat. You can think of heat as a "reactant." • Equation: $\text{Heat} + N_2O_4(g) \rightleftharpoons 2NO_2(g)$
• Effect of Temperature: • When you increase the temperature of an endothermic reaction, the system tries to use up the extra heat by shifting the equilibrium to the right (towards the products). • Because the concentration of products ($[NO_2]$) increases and the concentration of reactants ($[N_2O_4]$) decreases, the equilibrium constant ($K$) increases.

Step-by-Step Solution

1. Identify the reaction type: The value $\Delta H = +ve$ indicates that the forward reaction is endothermic.
2. Apply the temperature change: Increasing the temperature adds heat to the system.
3. Predict the shift: According to Le Chatelier’s Principle, the system will shift to the right to consume the added heat.
4. Determine the effect on $K$: For endothermic reactions, an increase in temperature favors the forward reaction, leading to a higher concentration of products relative to reactants. Since $K = \frac{[\text{Products}]}{[\text{Reactants}]}$, an increase in the numerator and a decrease in the denominator results in a larger value for $K$.

Evaluating the Options

• A. Increase the value of the equilibrium constant: This is correct because the forward reaction is favored in an endothermic process when temperature rises.
• B. Decreases the value of the equilibrium constant: This would only happen if the reaction were exothermic ($\Delta H = -ve$).
• C. Increase in the reactant production: Incorrect; the equilibrium shifts to the right, producing more $NO_2$ (product), not reactants.
• D. Shift the equilibrium to the left: Incorrect; the equilibrium shifts to the right to absorb the excess heat.

Exam Tip

Remember the "Temperature-K Rule":

• Endothermic ($\Delta H > 0$): Temperature $\uparrow \implies K \uparrow$
• Exothermic ($\Delta H < 0$): Temperature $\uparrow \implies K \downarrow$ A simple way to remember this is that in endothermic reactions, heat is a "friend" to the products, so adding more heat helps the reaction move forward!