Question

$ {{K}_{p}} $ for the reaction, $ {{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g) $ is 0.157 atm at $ 27{}^\circ C $ and 1 atm pressure. $ {{K}_{c}} $ for this reaction is

• A. $ 6.37\times {{10}^{-3}} $
• B. $ 4.29\times {{10}^{-3}} $
• C. $ 5.62\times {{10}^{-3}} $
• D. $ 4.37\times {{10}^{-3}} $

Answer 1

Answer: (A)

$ \Delta {{n}_{g}}={{n}_{p}}-{{n}_{r}}=2-1=1, $ $ {{K}_{c}}=\frac{{{K}_{p}}}{{{(RT)}^{\Delta {{n}_{g}}}}} $ $ =\frac{(0.157\,atm)}{{{(0.0821\,\,L\,\,atm\,{{K}^{-1}}\,mo{{l}^{-1}}\,\times \,300\,K)}^{1}}} $ $ =6.37\times {{10}^{-3}}\,mol\,{{L}^{-1}} $