question_answer 1)

Find out the value ofx\[{{K}_{c}}\] for each of the following equilibria from the value of \[{{K}_{p}}\]: (i) \[2NOCl(g)\underset{{}}{\leftrightarrows}2NO(g)+C{{l}_{2}}(g);\]\[{{K}_{p}}=1.8\times {{10}^{-2}}\] at 500 K (ii) \[CaC{{O}_{3}}(s)\underset{{}}{\leftrightarrows}CaO(s)+C{{O}_{2}}(g);\] \[{{K}_{p}}=167\] at 1073 K

Answer:

(i)\[2NOCl(g)\underset{{}}{\leftrightarrows}2NO(g)+C{{l}_{2}}(g)\] \[\Delta {{n}_{g}}=3-2=1\] \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\] \[{{K}_{c}}=\frac{{{K}_{p}}}{{{(RT)}^{\Delta {{n}_{g}}}}}=\frac{1.8\times {{10}^{-2}}}{(0.0821\times 500)}\] \[=4.384\times {{10}^{-4}}\,mol\,{{L}^{-1}}\] (ii) \[CaC{{O}_{3}}(s)\underset{{}}{\leftrightarrows}CaO(s)+C{{O}_{2}}(g)\] \[\Delta {{n}_{g}}=1-0=1\] \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\] \[{{K}_{c}}=\frac{{{K}_{p}}}{{{(RT)}^{\Delta {{n}_{g}}}}}\] \[=\frac{167}{{{(0.0821\times 1073)}^{3}}}\] \[=1.895\,mol\,{{L}^{-1}}\]