question_answer 4)

The value of \[{{K}_{c}}=4.24\]at 800 K for the reaction: \[CO(g)+{{H}_{2}}O(g)\rightleftharpoons C{{O}_{2}}(g)+{{H}_{2}}(g)\] Calculate equilibrium concentrations of \[C{{O}_{2}}(g),{{H}_{2}}(g)\]\[CO(g)\]and \[{{H}_{2}}O(g)\]at 800K, if only CO and \[{{H}_{2}}O\] are present initially at concentrations of 0.1 M each.

Answer:

The given reaction is: \[\underset{{{t}_{eq}}}{\overset{{}}{\mathop{{{t}_{0}}}}}\,\underset{\underset{(0.1-x)}{\mathop{0.1}}\,}{\mathop{CO(g)}}\,+\underset{\underset{(0.1-x)}{\mathop{0.1}}\,}{\mathop{{{H}_{2}}O(g)}}\,\rightleftharpoons \underset{\begin{smallmatrix} 0 \\ x \end{smallmatrix}}{\mathop{C{{O}_{2}}(g)}}\,+\underset{\underset{x}{\mathop{0}}\,}{\mathop{{{H}_{2}}(g)}}\,\] \[{{K}_{c}}=\frac{[C{{O}_{2}}][{{H}_{2}}]}{[CO][{{H}_{2}}O]}\] \[4.24=\frac{x\times x}{(0.1-x)(0.1-x)}\] \[\sqrt{4.24}=\frac{x}{0.1-x}\] \[2.06=\frac{x}{0.1-x}\] \[2.06(0.1-x)=x\] \[x=0.067\] \[\therefore \] \[[C{{O}_{2}}]=[{{H}_{2}}]=0.06M\] \[[C{{O}_{2}}]=[{{H}_{2}}O]=0.1-0.067\] \[=0.033M\]