91Ó°ÊÓ

(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 \(Q\_\left\{ p \right\} p\)