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(A) The equilibrium constant $(K_c)$ depends on the strength of the chemical bonds in the reactants and products.In reaction $i$,the $N-H$ bond is bro

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The equilibrium constant for reaction $i$ is greater than that for reaction $ii$.

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(A) The equilibrium constant $(K_c)$ depends on the strength of the chemical bonds in the reactants and products.In reaction $i$,the $N-H$ bond is broken,while in reaction $ii$,the $N-D$ bond is broken.The $N-D$ bond is stronger than the $N-H$ bond due to the higher mass of deuterium $(D)$,which results in a lower zero-point energy.Because the $N-D$ bond is stronger,it is more difficult to break than the $N-H$ bond.Consequently,the equilibrium constant for the reaction involving $ND_3$ $(K_{c,ii})$ is smaller than the equilibrium constant for the reaction involving $NH_3$ $(K_{c,i})$.Therefore,$K_{c,i} > K_{c,ii}$.

Consider the following two equilibrium reactions:
$i$. $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$
$ii$. $2ND_{3(g)} \rightleftharpoons N_{2(g)} + 3D_{2(g)}$
What is the difference in their equilibrium constants $(K_c)$?

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Consider the following two equilibrium reactions:$i$. $2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}$$ii$. $2ND_{3(g)} \rightleftharpoons N_{2(g)} + 3D_{2(g)}$What is the difference in their equilibrium constants $(K_c)$?