Question

For the reaction $ Fe{{O}_{(s)}}+C{{O}_{(g)}}F{{e}_{(s)}}+C{{O}_{2}}_{(g)},\,\,{{K}_{C}}=5.0. $ If the concentration of CO at equilibrium is $ 2.5\times {{10}^{-2}}mol/litre $ at $ {{100}^{o}}C $ . Then the equilibrium concentration of $ C{{O}_{2}} $ will be:

• A. $ 1.25\times {{10}^{-4}}m/L $
• B. $ 12.5\times {{10}^{-2}}m/L $
• C. $ 2.50\times {{10}^{-3}}m/L $
• D. $ 2.50\times {{10}^{-4}}m/L $

Answer 1

Answer: (B)

$ Fe{{O}_{(s)}}+C{{O}_{(g)}}\xrightarrow{{}}F{{e}_{(s)}}+C{{O}_{2(g)}} $ At equilibrium $ {{K}_{C}}=\frac{[C{{O}_{2}}]}{[CO]} $ given $ [CO]=2.5\times {{10}^{-2}}mol/litre,\,\,{{K}_{C}}=5.0 $ $ 5=\frac{[C{{O}_{2}}]}{2.5\times {{10}^{-2}}} $ $ [C{{O}_{2}}]=12.5\times {{10}^{-2}}mol/litre $