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The reaction, Xe(excess) + F2 to XeF2 is...

The reaction, Xe(excess) `+ F2 to XeF_2` is conducted at 

A

573 K, 16-70 bar

B

273 K, 10 bar

C

673 K, 1 bar

D

873 K, 7 bar

Text Solution

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The correct Answer is:
To solve the question regarding the reaction of xenon (Xe) with fluorine (F2) to form xenon difluoride (XeF2), we will analyze the conditions under which this reaction occurs. ### Step-by-Step Solution: 1. **Identify the Reaction**: The reaction given is: \[ \text{Xe} + \text{F}_2 \rightarrow \text{XeF}_2 \] Here, xenon reacts with fluorine to produce xenon difluoride. 2. **Understand the Conditions for the Reaction**: The formation of xenon difluoride typically occurs under specific conditions of temperature and pressure. 3. **Consider the Options Provided**: The options given for the temperature and pressure conditions are: - 573 K, 16 to 70 bar - 273 K, 10 bar - 673 K, 1 bar - 873 K, 7 bar 4. **Determine the Right Conditions**: - The reaction between xenon and fluorine to form XeF2 generally occurs at high temperatures. - The formation of higher fluorides (XeF4 and XeF6) can occur at even higher temperatures and pressures. - Based on chemical literature, the synthesis of XeF2 is typically conducted at temperatures around 673 K and pressures around 1 bar. 5. **Select the Correct Option**: Among the options provided, the one that matches the typical conditions for the formation of xenon difluoride is: - **673 K and 1 bar**. ### Final Answer: The reaction \( \text{Xe} + \text{F}_2 \rightarrow \text{XeF}_2 \) is conducted at **673 K and 1 bar pressure**. ---
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