91Ó°ÊÓ

The reaction quotient, abbreviated as \({Q_c}\), is a measure of the relative concentration of reactants and products present in a chemical reaction at a specific point in 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:

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.

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.

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

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

Where

\[\begin{array}{}{\rm{pA}} = {\rm {partial pressure of gaseous reactant}}\; {\rm{A}}\\{\rm{pB}} = {\rm{partial pressure of gaseous reactant}}\;{\rm{B}}\\{\rm{pC}} = {\rm{partial pressure of gaseous product}}\;{\rm{C}}\\{\rm{pD}} = {\rm{partial pressure of gaseous product}}\;{\rm{D}}\end{array}\]

(a)

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

\({{\rm{Q}}_{\rm{c}}} = \frac{{\left( {{{\rm{N}}_2}\mid {{\left( {{{\rm{H}}_2}} \right)}^3}} \right.}}{{{{\left( {{\rm{N}}{{\rm{H}}_3}} \right)}^2}}}\)

Given,

\(\begin{aligned}{}{{\bf{K}}_{\rm{c}}} = 17\\({\rm{N}}{{\rm{H}}_3}) = 0.50\;M\\({{\rm{N}}_2}) = 0.15\;M\\({{\rm{H}}_2}) = 0.12\;M\\{Q_c} = \frac{{(0.15{\rm{M}}) \times {{(0.12{\rm{M}})}^3}}}{{{{(0.50{\rm{M}})}^2}}}\\ = 0.00864\end{aligned}\)

Therefore, \({Q_C}c\) will proceed in forward direction.

(b)

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

\({{\rm{Q}}_{\rm{p}}} = \frac{{\left( {{\rm{p}}{{\rm{N}}_2}} \right){{\left( {{\rm{p}}{{\rm{H}}_2}} \right)}^5}}}{{{{\left( {{\rm{pN}}{{\rm{H}}_3}} \right)}^2}}}\)

Given,

\(\begin{aligned}{}{{\rm{K}}_{\rm{p}}} = 6.8 \times {10^4}\\{\rm{pN}}{{\rm{H}}_3} = 2.00\;{\rm{atm,}}\\{\rm{p}}{{\rm{N}}_2} = 10.00\;{\rm{atm,}}\\{\rm{p}}{{\rm{H}}_2} = 10.00\;{\rm{atm,}}\\{{\rm{Q}}_{\rm{p}}} = \frac{{(10.00\;{\rm{atm}}){{(10.00\;{\rm{atm}})}^3}}}{{{{(2.00\;{\rm{atm}})}^2}}}\\ = 2500\end{aligned}\)

Therfore, \({Q_p} > {K_p}\), the reaction will proceed in reverse direction.

(c)

For reaction \(2{\rm{S}}{{\rm{O}}_3}(\;g)\rightleftharpoons 2{\rm{S}}{{\rm{O}}_2}(\;g) + {{\rm{O}}_2}(\;g)\)

\({{\rm{Q}}_{\rm{c}}} = \frac{{\left( {{{\left. {{\rm{S}}{{\rm{O}}_2}} \right|}^2}\left( {{{\rm{O}}_2}} \right)} \right.}}{{{{\left( {{\rm{S}}{{\rm{O}}_3}} \right)}^2}}}\)

Given,

\(\begin{aligned}{}{{\rm{K}}_{\rm{c}}} = 0.230\\\left( {{\rm{S}}{{\rm{O}}_3}} \right) = 2.00\;{\rm{M}}\\\left( {{\rm{S}}{{\rm{O}}_2}} \right) = 2.00\;{\rm{M}}\\\left( {{{\rm{O}}_2}} \right) = 2.00\;{\rm{M}}\\{{\rm{Q}}_{\rm{c}}} = \frac{{{{(2.00\;{\rm{M}})}^2} \times (2.00\;{\rm{M}})}}{{{{(2.00\;{\rm{M}})}^2}}} = 2.00\end{aligned}\)

Therefore, \({Q_c} > {K_c}\) will proceed in reverse direction.

(d)

For reaction \(2{\rm{S}}{{\rm{O}}_3}(\;g)\rightleftharpoons 2{\rm{S}}{{\rm{O}}_2}(\;g) + {{\rm{O}}_2}(\;g)\)

\({{\rm{Q}}_{\rm{p}}} = \frac{{{{\left( {{\rm{pS}}{{\rm{O}}_2}} \right)}^2}\left( {{\rm{p}}{{\rm{O}}_2}} \right)}}{{{{\left( {{\rm{pS}}{{\rm{O}}_3}} \right)}^2}}}\)

Given,

\(\begin{aligned}{}{{\rm{K}}_{\rm{p}}} = 6.5\\{\rm{pS}}{{\rm{O}}_3} = 0\;{\rm{atm}}\\{\rm{pS}}{{\rm{O}}_2} = 1.00\;{\rm{atm}}\\{\rm{p}}{{\rm{O}}_2} = 1.130\;{\rm{atm}}\\{{\rm{Q}}_{\rm{p}}} = \frac{{{{(1.00\;{\rm{atm}})}^2}(1.130\;{\rm{atm}})}}{{{{(0\;{\rm{atm}})}^2}}}\\ = \infty \end{aligned}\)

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

(e)

For reaction \(2{\rm{NO}}(g) + {\rm{C}}{{\rm{l}}_2}(g) \rightleftharpoons 2{\rm{NOCl}}(g)\)

\({{\rm{Q}}_{\rm{c}}} = \frac{{{{({\rm{NOCl}})}^2}}}{{\left( {{\rm{N}}{{\rm{O}}^2}\left( {{\rm{C}}{{\rm{l}}_2}} \right)} \right.}}\)

Given,

\(\begin{aligned}{}{{\rm{K}}_{\rm{c}}} = 2.5 \times {10^3}\\({\rm{NO}}) = 1.00\,{\rm{M}}\\({\rm{Cl}}2) = 1.00\,{\rm{M}}\\({\rm{NOCl}}) = 0\,{\rm{M}}\\{{\rm{Q}}_{\rm{c}}} = \frac{{{{(0\,{\rm{M}})}^2}}}{{{{(1.00\,{\rm{M}})}^2}(1.00\,{\rm{M}})}}\\ = 0\end{aligned}\)

Therefore, \({Q_c}c\)will proceed in forward direction.

(f)

For reaction \({{\rm{N}}_2}(\;g) + {{\rm{O}}_2}(\;g) \rightleftharpoons 2{\rm{NO}}(g)\)

\({{\rm{Q}}_{\rm{p}}} = \frac{{{{({\rm{pNO}})}^2}}}{{\left( {{\rm{p}}{{\rm{N}}_2}} \right)({\rm{p}}{{\rm{O}}_{\rm{2}}}{\rm{)}}}}\)

Given,

\(\begin{aligned}{}{{\rm{K}}_{\rm{p}}} = 0.050\\{\rm{pNO}} = 1.00\;{\rm{atm,}}\\{\rm{p}}{{\rm{N}}_2} = 0.100\;{\rm{atm}},\\{\rm{p}}{{\rm{O}}_2} = 0.200\;{\rm{atm,}}\\{{\rm{Q}}_{\rm{p}}} = \frac{{{{(1.00\;{\rm{atm}})}^2}}}{{(0.100\;{\rm{atm}})(0.200\;{\rm{atm}})}}\\ = 50\end{aligned}\)

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