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  4. Xenon reacts with fluorine at 873 K and 7 bar to form XeF 4. In this reaction, the ratio of xenon and fluorine required is

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Q. Xenon reacts with fluorine at $873\, K$ and $7$ bar to form $XeF _{4}$. In this reaction, the ratio of xenon and fluorine required is

TS EAMCET 2016

Solution:

$XeF _{4}$ is obtained by heating a mixture of xenon and fluorine in the molar ratio of $1: 5$ at $873\, K$ and $7$ bar pressure in an enclosed nickel vessel for a few hours. The reaction proceeds as

$Xe (g)+2 F _{2}(g) \ce{->[{873 K }][{7 \text{bar}}]} XeF _{4}$

The extra fluorine taken, increases the production.