A) \[\frac{{{K}_{C}}}{{{K}_{P}}}={{(RT)}^{\Delta {{n}_{g}}}}\]
B) \[\frac{{{K}_{p}}}{{{K}_{x}}}={{(p)}^{\Delta {{n}_{g}}}}\]
C) \[\frac{{{K}_{C}}}{{{K}_{x}}}={{(p)}^{-\Delta {{n}_{g}}}}\]
D) \[\frac{{{K}_{C}}}{{{K}_{x}}}={{\left( \frac{p}{RT} \right)}^{\Delta {{n}_{g}}}}\]
Correct Answer: B
Solution :
Since, \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\] \[\therefore \] \[\frac{{{K}_{p}}}{{{K}_{c}}}={{(RT)}^{\Delta {{n}_{g}}}}\] and \[{{K}_{p}}={{K}_{x}}{{(p)}^{\Delta {{n}_{g}}}}\] \[\therefore \] \[\frac{{{K}_{p}}}{{{K}_{x}}}={{(p)}^{\Delta {{n}_{g}}}}\] In terms of mole fraction, ie \[{{K}_{x}}\] \[{{K}_{x}}=\frac{[{{X}_{c}}]_{eq}^{n1}.[{{X}_{\circ }}]_{eg...}^{n2}}{{{[{{X}_{A}}]}^{n1}}[{{X}_{B}}]_{eq...}^{m2}}\] If \[\Delta n=0\] then \[{{K}_{p}}={{K}_{c}}={{K}_{x}}\] \[{{K}_{x}}\]differs from \[{{K}_{p}}\] and \[{{K}_{c}}\] in that its value depends uon the P and V of the system if \[\Delta n\ne 0,\] while under such condition \[{{K}_{p}}\] and \[{{K}_{c}}\] remain constant and vary with temperature and that is why, \[{{K}_{x}}\] is not used in connection with gas relations.
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