The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction) in which each system will proceed to leach equilibrium.

(a)

Because of the reaction,\(2N{H_3}(g) \rightleftharpoons {N_2}(g) + 3{H_2}(g)\)

\({Q_c} = \frac{{\left| {{N_2}} \right|{{\left| {{H_2}} \right|}^3}}}{{{{\left| {N{H_3}} \right|}^2}}}\)

Given,

\(\begin{aligned}{}{K_c} = 17;\left( {N{H_3}} \right) = 0.20M,\left( {{N_2}} \right) = 1.00M,\left( {{H_2}} \right) = 1.00M\\{Q_c} = \frac{{(1.00M) \times {{(1.00M)}^3}}}{{{{(0.20M)}^2}}} = 25\end{aligned}\)

As a result, the reaction will go in the opposite direction \({Q_C} > {K_c}\)

(b)

Because of the reaction\(2N{H_3}(g)\rightleftharpoons {N_2}(g) + 3{H_2}(g)1.0\;atm\)

\({Q_p} = \frac{{\left( {p{N_2}} \right){{\left( {p{H_2}} \right)}^3}}}{{{{\left( {pN{H_3}} \right)}^2}}}\)

Given,

\(\begin{aligned}{}{{\rm{K}}_{\rm{p}}} = 6.8 \times {10^4};{\rm{pN}}{{\rm{H}}_3} = 3.0\;{\rm{atm}},{\rm{p}}{{\rm{N}}_2} = 2.0\;{\rm{atm}},{\rm{p}}{{\rm{H}}_2} = 1.0\;{\rm{atm}}\\{{\rm{Q}}_{\rm{p}}} = \frac{{(2.0\;{\rm{atm}}){{(1.0\;{\rm{atm}})}^3}}}{{{{(3.0\;{\rm{atm}})}^2}}} = 0.22\end{aligned}\)

As a result, the response will move forward \(Q\_\left\{ p \right\} p\)

(c)

Because of the reaction,\(2S{O_3}(g)\rightleftharpoons 2S{O_2}(g) + {O_2}(g)\)

\({Q_c} = \frac{{\left. {{{\left| {S{O_2}} \right|}^2}\mid {O_2}} \right)}}{{\left( {{{\left. {S{O_3}} \right|}^2}} \right.}}\)

Given,

\(\begin{aligned}{l}{K_c} = 0.230;\left( {S{O_3}} \right) = 0.00M,\left( {S{O_2}} \right) = 1.00M,\left( {{O_2}} \right) = 1.00M\\{Q_c} = \frac{{{{(1.00M)}^2} \times (1.00M)}}{{{{(0.00M)}^2}}} = \propto \end{aligned}\)

As a result, the reaction will go in the opposite direction, \({Q_C} > {K_C}\)

(d)

Because of the reaction

\(2S{O_3}(g) \rightleftharpoons 2S{O_2}(g) + {O_2}(g)1.00\;atm\)

\({Q_p} = \frac{{{{\left( {pS{O_2}} \right)}^2}\left( {p{O_2}} \right)}}{{{{\left( {pS{O_3}} \right)}^2}}}\)

Given,

\(\begin{aligned}{l}{K_p} = 16.5;pS{O_3} = 1.00\;atm,pS{O_2} = 1.00\;atm,p{O_2} = 1.0\;atm\\{Q_p} = \frac{{{{(1.00\;atm)}^2}(1.00\;atm)}}{{{{(1.00\;atm)}^2}}} = 1.0\end{aligned}\)

As a result, the response will move forward, \(Q\_\left\{ p \right\} p\)

(e)

Because of the reaction\(2NO(g) + C{l_2}(g) \rightleftharpoons 2NOCl(g)\)

\({Q_c} = \frac{{{{(NOCl)}^2}}}{{\left( {N{O^2}\left( {C{l_2}} \right)} \right.}}\)

Given,

\(\begin{aligned}{}{K_c} = 4.6 \times 1{0^4};(NO) = 1.00M,\left( {C{l_2}} \right) = 1.00M,(NOCl) = 0M\\{Q_c} = \frac{{{{(0M)}^2}}}{{\left( {1.00{M^2}(1.00M)} \right.}} = 0\end{aligned}\)

As a result, the response will move forward \(Q\_\left\{ C \right\} c\).

(f)

Because of the reaction \({N_2}(g) + {O_2}(g) \rightleftharpoons 2NO(g)\)

\({Q_p} = \frac{{\left( {pN{O^2}} \right.}}{{\left( {p{N_2}} \right)\left( {p{O_2}} \right)}}\)

Given,

\(\begin{aligned}{l}{K_p} = 0.050;pNO = 10.0\;atm,p{N_2} = 5\;atm,p{O_2} = 5\;atm\\{Q_p} = \frac{{{{(10.0\;atm)}^2}}}{{(5\;atm)(5\;atm)}} = 4\end{aligned}\)

As a result, the response will move forward \(Q\_\left\{ p \right\} p\)