In the gaseous equilibrium
$H_{2} X_{2} (g) + h e a t ⇌ 2 H X (g)$ , the formation of $H X$ will be favoured by

High pressure and low temperature

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High temperature and low pressure

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Low temperature and low pressure

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High temperature and high pressure

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Solution

The correct option is B High temperature and low pressure
$H_{2} X_{2} (g) + h e a t ⇌ 2 H X (g)$ . Reaction is endothermic and number of moles increasing in forward direction so according to Le chatelier's principle for formation of HX, Temperature of the reaction should be high and pressure should be low since decrease in pressure favors the side that has more number of moles.

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