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

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High temperature and low pressure

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(C) The dissociation of molecular hydrogen into atomic hydrogen is represented by the equation: $H_{2(g)} ightleftharpoons 2H_{(g)}$.
This reaction involves the breaking of a chemical bond,which is an endothermic process $(\Delta H > 0)$.
According to Le Chatelier's principle,for an endothermic reaction,an increase in temperature favors the forward reaction.
Furthermore,the number of moles of gas increases from $1 	ext{ mole}$ on the reactant side to $2 	ext{ moles}$ on the product side.
According to Le Chatelier's principle,a decrease in pressure favors the side with a greater number of moles of gas.
Therefore,high temperature and low pressure are the ideal conditions for the dissociation of $H_2$ into $H$.

Under which of the following conditions of temperature and pressure will molecular hydrogen $(H_2)$ dissociate into atomic hydrogen $(H)$?

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