question_answer 1)

The following concentrations were obtained for the formation of\[N{{H}_{3}}\] from \[{{N}_{2}}\] and \[{{H}_{2}}\] at equilibrium at 500 K. \[[{{N}_{2}}]=1.5\times {{10}^{-2}}M,[{{H}_{2}}]=3\times {{10}^{-2}}M\]and\[[N{{H}_{3}}]=1.2\times {{10}^{-2}}M.\]Calculate the equilibrium constant.

Answer:

The reaction is: \[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)\] \[{{K}_{c}}=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]{{[{{H}_{2}}]}^{3}}}\] \[=\frac{{{[1.2\times {{10}^{-2}}]}^{2}}}{[1.5\times {{10}^{-2}}]{{[13\times {{10}^{-2}}]}^{3}}}\] \[=3.55\times {{10}^{2}}mo{{l}^{-2}}{{L}^{2}}\]