The following reaction has \({K_P} = 4.50 \times {10^{ - 5}}\) at \(720\;{\rm{K}}\).

\({{\rm{N}}_2}(g) + 3{{\rm{H}}_2}(g) \rightleftharpoons 2{\rm{N}}{{\rm{H}}_3}(g)\)

If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium?

\(P\left( {{\rm{N}}{{\rm{H}}_3}} \right) = 93\;{\rm{atm}},\;P\left( {\;{{\rm{N}}_2}} \right) = 48\;{\rm{atm}},\;{\rm{and}}\;P\left( {{{\rm{H}}_2}} \right) = 52\)

The reaction quotient also symbolized as\({Q_c}\), is a measurement of the relative concentrations of reactants and products present in a chemical process at a given time.When the system is in equilibrium, however, \({{\rm{K}}_{\rm{c}}}\) or, the equilibrium constant, is used to calculate the specific value of the reaction quotient. In order to determine the direction of a reaction at equilibrium, compare \({Q_c}\)with \({{\rm{K}}_{\rm{c}}}\). When forecasting the direction of a reaction, there are three factors to consider.

The partial pressure of the gases present at equilibrium define the reaction quotient, \({Q_p}\), and the equilibrium constant, \({K_p}\), for a gaseous system.

The equilibrium constant \({{\rm{K}}_{\rm{c}}}\)is used to calculate the specific value of the reaction quotient when the system is at equilibrium. Compare with to figure out which way a reaction is going while it's at equilibrium. There are three things to consider when predicting the direction of a reaction.

The reaction quotient, and the equilibrium constant, for a gaseous system are defined by the partial pressures of the gases present at equilibrium.

Thus, the reversible reaction of the form, \(aA + bB \rightleftharpoons cC + dD\)

\({Q_p} = \frac{{{{(pC)}^c}{{(pD)}^d}}}{{{{(pA)}^a}{{(pB)}^b}}}\)

Where

\(\begin{aligned}{c}pA = {\rm{partial pressure of gaseous reactant}}\;{\rm{A}}\\pB = {\rm{partial pressure of gaseous reactant}}\;{\rm{B}}\\pC = {\rm{partial pressure of gaseous product}}\;{\rm{C}}\\pD = {\rm{partial pressure of gaseous product}}\;{\rm{D}}\end{aligned}\)

When the numerator, or the amount of product existing at any one time, is more than the denominator, or the amount of reactant, this condition exists. When surplus product is present, the system tends to go in the direction of reducing the excess product/s by returning the reactant/s to achieve equilibrium (according to Le Chatlier's principle). As a result, the reaction occurs in reverse.

When the amount of reactant/s and product/s are constant, the reaction is already at equilibrium. This is because the forward reaction rate is equal to the reverse reaction rate, and the system has no inclination to create any more product/s or reactant/s. As a result, there is no change in the direction of the response.

When the numerator, or the amount of product existing at any one time, is less than the denominator, or the amount of reactant, this condition exists. When surplus reactant(s) are present, the system tends to proceed in the direction of reducing the excess reactant(s) by returning the product(s) in order to achieve equilibrium (as per Le Chatlier's principle). As a result, the reaction is moving forward.

For the reaction \({{\rm{N}}_2}(g) + 3{{\rm{H}}_2}(g) \rightleftharpoons 2{\rm{N}}{{\rm{H}}_3}(g)\)

\({Q_p} = \frac{{{{\left( {p{\rm{N}}{{\rm{H}}_3}} \right)}^2}}}{{\left( {p\;{{\rm{N}}_2}} \right){{\left( {p{{\rm{H}}_2}} \right)}^3}}}\)

Given,

$

\(\begin{aligned}{c}{{\rm{K}}_{\rm{p}}} = 4.50 \times {10^5};\\{\rm{pN}}{{\rm{H}}_3} = 93\;{\rm{atm}},\\{\rm{p}}{{\rm{N}}_2} = 48\;{\rm{atm}},\\{\rm{p}}{{\rm{H}}_2} = 52\;{\rm{atm}}\\{{\rm{Q}}_{\rm{p}}} = \frac{{{{(93\;{\rm{atm}})}^2}}}{{(48\;{\rm{atm}}){{(52\;{\rm{atm}})}^3}}}\\ = 1.3 \times {10^{ - 3}}\end{aligned}\)

Therefore, \({Q_p} > {K_p}\) will proceed in reverse direction.