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 \[{{p}_{CO}}=2x=0.66bar\]
\[{{p}_{C{{O}_{2}}}}=(0.48-x)\]
\[=(0.48-0.33)=0.15bar\]
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