question_answer 1)

A mixture of 1.57 mole of \[{{N}_{2}}\] 1.92 mole of \[{{H}_{2}}\] and 8.13mole of\[N{{H}_{3}}\] is introduced into a 20 L reaction vessel at 500K. At this temperature, the equilibrium constant \[{{K}_{c}}\] for the reaction: \[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)\]   is\[1.7\times {{10}^{2}}\].Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?

Answer:

The reaction is,\[\underset{\begin{smallmatrix} Concentration \\ of\,reacting \\ components \end{smallmatrix}}{\overset{{}}{\mathop{{}}}}\,\underset{\left[ \frac{1.57}{20} \right]}{\mathop{{{N}_{2}}(g)}}\,+\underset{\left[ \frac{1.92}{20} \right]}{\mathop{3{{H}_{2}}(g)}}\,\rightleftharpoons \underset{\left[ \frac{8.13}{20} \right]}{\mathop{2N{{H}_{3}}(g){{K}_{c}}}}\,=1.7\times {{10}^{2}}\] \[{{Q}_{c}}=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]{{[{{H}_{2}}]}^{3}}}=\frac{{{\left( \frac{8.13}{20} \right)}^{2}}}{\left[ \frac{1.57}{20} \right]{{\left[ \frac{1.92}{20} \right]}^{3}}}\] \[\begin{align} & =\frac{{{(8.13)}^{2}}\times {{(20)}^{2}}}{1.57\times {{(1.92)}^{3}}}=2.3792\times {{10}^{3}} \\ & \\ \end{align}\] \[{{Q}_{c}}>{{K}_{c}}\] The reaction is not at equilibrium. The net reaction will be in backward direction.