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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 is end

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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: $H_2(g) \rightleftharpoons 2H(g)$.This reaction is endothermic $( \Delta H > 0 )$ and involves an increase in the number of moles of gas $( \Delta n_g = 2 - 1 = 1 )$.According to Le Chatelier's principle:$1$. Since the reaction is endothermic,a high temperature will shift the equilibrium to the right (product side).$2$. Since the reaction involves an increase in the number of moles of gas,a low pressure will shift the equilibrium to the right (product side).Therefore,the formation of atomic hydrogen is favoured at high temperature and low pressure.

Under what conditions of temperature and pressure will the formation of atomic hydrogen from molecular hydrogen be favoured most?

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