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(C) The dissociation of molecular hydrogen is represented by the equation: $H_{2(g)} \rightleftharpoons 2H_{(g)}$.
This reaction is endothermic $(\De

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High temperature and low pressure

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(C) The dissociation of molecular hydrogen is represented by the equation: $H_{2(g)} \rightleftharpoons 2H_{(g)}$.
This reaction is endothermic $(\Delta H > 0)$ because energy is required to break the $H-H$ bond.
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature favors the forward reaction.
Regarding pressure,the number of moles of gas increases from $1$ mole on the reactant side to $2$ moles on the product side.
According to Le Chatelier's principle,a decrease in pressure favors the side with more moles of gas,which is the product side.
Therefore,high temperature and low pressure favor the dissociation of molecular hydrogen.

Under which conditions of temperature and pressure is the dissociation of molecular hydrogen into atomic hydrogen favored?

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