question_answer 6)

The value of K p for the reaction, \[C{{O}_{2}}(g)+C(s)\rightleftharpoons 2CO(g)\]is 3 at 1000K. If initially\[{{p}_{C{{O}_{2}}}}=0.48\]bar and \[{{p}_{CO}}=0\]bar and pure graphite is present, calculate the equilibrium partial pressures of CO and\[C{{O}_{2}}\].

Answer:

The given reaction is: \[\underset{{{t}_{eq}}}{\overset{{}}{\mathop{{{t}_{0}}}}}\,\underset{\underset{(0.48-x)}{\mathop{0.48}}\,}{\mathop{CO(g)}}\,+\underset{{}}{\mathop{{{H}_{2}}O(g)}}\,\rightleftharpoons \underset{{}}{\mathop{C{{O}_{2}}(g)}}\,+\underset{\underset{2x}{\mathop{0}}\,}{\mathop{{{H}_{2}}(g)}}\,\] \[{{K}_{p}}=\frac{{{[{{p}_{CO}}]}^{2}}}{[{{p}_{C{{O}_{2}}}}]}=\frac{{{(2x)}^{2}}}{0.48-x}=3\] \[4{{x}^{2}}+3x-1.44=0\] \[x=\frac{-3\pm \sqrt{9+4\times 4\times 1.44}}{2\times 4}=\frac{-3\pm \sqrt{32.04}}{8}\] \[x=\frac{-3\pm 5.66}{8}=0.33\] (taking + ve sign) At equilibrium \[{{p}_{CO}}=2x=0.66bar\] \[{{p}_{C{{O}_{2}}}}=(0.48-x)\] \[=(0.48-0.33)=0.15bar\]