At room temperature, the equilibrium constant for the reaction P + Q `hArr` R + S was calculated to be `4.32` . At `425^(@)C` the equilibrium constant became `1.24xx10^(-2)` . This indicates that the reaction

Options

A. is exothermic
B. is endothermic
C. is difficult to predict
D. no relation between `DeltaH "and" K`

Explanation

To determine whether the reaction \( P + Q \rightleftharpoons R + S \) is exothermic or endothermic based on the given equilibrium constants at different temperatures, we can follow these steps:

### Step 1: Identify the given data
- At room temperature (approximately \( 298 \, \text{K} \)), the equilibrium constant \( K_1 = 4.32 \).
- At \( 425^\circ C \) (which is \( 698 \, \text{K} \)), the equilibrium constant \( K_2 = 1.24 \times 10^{-2} \).

### Step 2: Analyze the change in equilibrium constant
We observe that:
- \( K_1 = 4.32 \) (at \( 298 \, \text{K} \))
- \( K_2 = 1.24 \times 10^{-2} \) (at \( 698 \, \text{K} \))

### Step 3: Determine the effect of temperature on the equilibrium constant
The equilibrium constant decreases as the temperature increases:
- From \( K_1 \) to \( K_2 \), we see a significant drop in the value of the equilibrium constant.

### Step 4: Apply Le Chatelier's Principle
According to Le Chatelier's Principle:
- If the equilibrium constant decreases with an increase in temperature, the reaction favors the endothermic direction (the reverse reaction in this case).

### Step 5: Conclude the nature of the reaction
Since the reverse reaction is favored at higher temperatures, it indicates that the forward reaction (which is \( P + Q \rightleftharpoons R + S \)) must be exothermic. This is because:
- An exothermic reaction releases heat, and increasing the temperature shifts the equilibrium towards the reactants (backward reaction).

### Final Answer
Thus, the reaction \( P + Q \rightleftharpoons R + S \) is an **exothermic reaction**.

---