Question

# The change in Pressure will not affect the equilibrium constant for A. ${N_2}\left( g \right) + 3{H_2}\left( g \right) \to \;2N{H_3}$B.$PC{l_5}\;\left( s \right) \to \;PC{l_3}\left( s \right) + C{l_2}(g)$C.$\;{H_2}\left( g \right) + {I_2}\left( g \right) \to \;2HI$D.All of the above

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Hint:
Apply Le Chatlelier’s Principle that governs the nature effect of Pressure, Temperature, and Concentration on equilibrium of the reaction. The change in pressure will not affect the equilibrium of those reactions that contain no difference in the number of moles of products and reactants.

When the number of molecules or atoms dissociating and going to products is equal to the number of molecules or atoms dissociating and going to reactants is equal we attain a state of equilibrium. When temperature is changed say for example is increased then more number of reactants will dissociate giving more number of product molecules. This change in temperature has altered the equilibrium. Similarly, if Pressure of the system containing gaseous reactants or products is changed then equilibrium may be hampered but condition is the number of moles of reactant should not be equal to the number of moles of products. If the number of moles of reactants is equal to the number of moles of product, then a change in pressure does not alter equilibrium.
As we look at the above equations.
-In (A) and (B) the total number of moles of the reactant is not equal to the total number of moles of product.
-But if we look at (C) reactants have a total of moles and products also have a total of 2 moles. So change in Pressure will not affect reaction (c). Conceptually thinking if pressure is increased then we have an equal number of moles to revert back thus no change happens.
${H_2} + {I_2} \to 2HI$
Number of moles of product = 2
Number of moles of reactants $= {\text{ }}1 + 1 = 2$
Total change $= {\text{ }}2 - 2{\text{ }} = {\text{ }}0$

Hence, option C is correct.