The equilibrium constants for the reaction, `A_2hArr2A` A at `500K and 700K` are `1xx10^(-10)` and `1xx10^(-5)`. The given reaction is

Options

A. exothermic
B. slow
C. endothermic
D. fast

Explanation

To solve the problem regarding the equilibrium constants for the reaction \( A_2 \rightleftharpoons 2A \) at two different temperatures, we can follow these steps:

### Step 1: Understand the Reaction and Equilibrium Constants
The reaction given is \( A_2 \rightleftharpoons 2A \). The equilibrium constant \( K_c \) for this reaction is defined as:
\[ K_c = \frac{[A]^2}{[A_2]} \]
where \([A]\) is the concentration of \( A \) and \([A_2]\) is the concentration of \( A_2 \).

### Step 2: Analyze the Given Data
We are provided with the following data:
- At \( 500 \, K \), \( K_c = 1 \times 10^{-10} \)
- At \( 700 \, K \), \( K_c = 1 \times 10^{-5} \)

### Step 3: Compare the Equilibrium Constants
Notice that as the temperature increases from \( 500 \, K \) to \( 700 \, K \), the value of \( K_c \) increases from \( 1 \times 10^{-10} \) to \( 1 \times 10^{-5} \). This indicates that the equilibrium position is shifting towards the products.

### Step 4: Determine the Nature of the Reaction
The increase in \( K_c \) with an increase in temperature suggests that the reaction is endothermic. In an endothermic reaction, heat is absorbed, and according to Le Chatelier's principle, increasing the temperature will favor the formation of products.

### Step 5: Conclusion
Since the equilibrium constant increases with temperature, we conclude that the reaction \( A_2 \rightleftharpoons 2A \) is endothermic.

### Final Answer
The given reaction is endothermic.

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