The equilibrium constant K for the reaction `2HI(g) hArr H_(2)(g)+I_(2)(g)` at room temperature is `2.85` and that at `698 K` is `1.4 xx10^(-2)`. This implies

Options

A. Exothermic
B. Endothermic
C. Exergonic
D. Unpredictable

Explanation

To determine the nature of the reaction based on the given equilibrium constants at two different temperatures, we can follow these steps:

### Step 1: Understand the Reaction
The reaction given is:
\[ 2 \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \]

### Step 2: Identify the Equilibrium Constants
We have two equilibrium constants:
- At room temperature (approximately 298 K), \( K = 2.85 \)
- At 698 K, \( K = 1.4 \times 10^{-2} \)

### Step 3: Analyze the Change in Equilibrium Constant with Temperature
The equilibrium constant \( K \) is dependent on temperature. For exothermic reactions, increasing the temperature typically decreases the value of \( K \), while for endothermic reactions, increasing the temperature increases the value of \( K \).

### Step 4: Compare the Values of \( K \)
- At 298 K, \( K = 2.85 \) (higher value)
- At 698 K, \( K = 1.4 \times 10^{-2} \) (much lower value)

Since the value of \( K \) decreases when the temperature is increased from 298 K to 698 K, this suggests that the reaction is exothermic.

### Conclusion
Since the equilibrium constant decreases with an increase in temperature, we conclude that the reaction is exothermic.

### Final Answer
The correct implication is that the reaction is **exothermic**.

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