Pure ammonia is placed in a vessel at a tempe | vedclass

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Question

(A) The dissociation reaction is $2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$.$K_p$ is a constant that depends only on temperature,not on pressure.A

Vedclass · Accepted Answer

$K_p$ does not change significantly with pressure.

Vedclass · Answer

(A) The dissociation reaction is $2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$.$K_p$ is a constant that depends only on temperature,not on pressure.According to Le Chatelier's principle,as pressure increases,the equilibrium shifts towards the side with fewer moles of gas (left side),so the degree of dissociation $(\alpha)$ decreases.From the stoichiometry of the reaction,for every $1$ mole of $N_2$ produced,$3$ moles of $H_2$ are produced,therefore the concentration of hydrogen is greater than that of nitrogen $([H_2] > [N_2])$.Thus,$K_p$ remains constant with pressure.

Pure ammonia is placed in a vessel at a temperature where its dissociation constant $(\alpha)$ is appreciable. At equilibrium,

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