Question

With an increase in temperature, the equilibrium constant for an endothermic reaction

• A. increases
• B. decreases
• C. increases linearly
• D. remains unaffected

Hint:

A simple mnemonic: For an endothermic reaction (\(\Delta H \gt 0\)), heat is a reactant. Adding heat (increasing T) pushes the reaction forward, so \(K\) increases. For an exothermic reaction (\(\Delta H \lt 0\)), heat is a product. Adding heat pushes the reaction backward, so \(K\) decreases.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to determine how the equilibrium constant (\(K\)) of an endothermic reaction changes when the temperature is increased. This can be understood using Le Chatelier's Principle or the Van't Hoff equation.

Step 2: Key Formula or Approach:

1. Le Chatelier's Principle: If a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress. For a temperature change, we can treat heat as a reactant or a product.

2. Van't Hoff Equation: This equation quantitatively relates the change in the equilibrium constant to the change in temperature: \[ \frac{d(\ln K)}{dT} = \frac{\Delta H^\circ}{RT^2} \] where \(\Delta H^\circ\) is the standard enthalpy change of the reaction.

Step 3: Detailed Explanation:

Using Le Chatelier's Principle:
An endothermic reaction is one that absorbs heat from the surroundings. We can write it schematically as: \[ \text{Reactants} + \text{Heat} \rightleftharpoons \text{Products} \] Here, heat is like a reactant. According to Le Chatelier's Principle, if we increase the temperature, we are "adding" heat to the system. The system will try to counteract this change by consuming the added heat. It does this by shifting the equilibrium to the right, favoring the formation of products. When the concentration of products increases and reactants decrease, the value of the equilibrium constant, \(K = \frac{[\text{Products}]}{[\text{Reactants}]}\), increases.

Using the Van't Hoff Equation:
For an endothermic reaction, the enthalpy change is positive (\(\Delta H^\circ \gt 0\)). Let's look at the Van't Hoff equation: \[ \frac{d(\ln K)}{dT} = \frac{\Delta H^\circ}{RT^2} \] Since \(\Delta H^\circ \gt 0\), \(R\) (gas constant) is positive, and \(T^2\) is always positive, the right side of the equation is positive. \[ \frac{d(\ln K)}{dT} \gt 0 \] This means that the slope of a plot of \(\ln K\) versus \(T\) is positive. In other words, as temperature (\(T\)) increases, \(\ln K\) increases, and therefore the equilibrium constant (\(K\)) itself increases. The relationship is not linear, but logarithmic.

Step 4: Final Answer:
For an endothermic reaction, increasing the temperature shifts the equilibrium towards the products, causing the equilibrium constant to increase.