Question
A vessel at $$1000 K$$ contains $$C{O_2}$$ with a pressure of 0.5 atm. Some of the $$C{O_2}$$ is converted into $$CO$$ on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of $$K$$ is :
A.
1.8 $$atm$$
B.
3 $$atm$$
C.
0.3 $$atm$$
D.
0.18 $$atm$$
Answer :
1.8 $$atm$$
Solution :
\[\underset{\begin{smallmatrix}
{{P}_{\text{initial}}}\,0.5\,atm \\
{{P}_{\text{final}}}\left( 0.5-x \right)\,atm
\end{smallmatrix}}{\mathop{C{{O}_{2}}+{{C}_{\left( \text{graphite} \right)}}}}\,\rightleftharpoons \underset{\begin{smallmatrix}
0 \\
2x\,atm
\end{smallmatrix}}{\mathop{2CO}}\,\]
$$\eqalign{
& {\text{Total }}P{\text{ at equilibrium }} \cr
& {\text{ = }}\,{\text{0}}{\text{.5}} - x + 2x = 0.5 + x\,atm \cr
& 0.8 = 0.5 + x \cr
& \therefore \,\,x = 0.8 - 0.5 = 0.3\,atm \cr
& {\text{Now}}\,\,{k_p} = \frac{{{{\left( {{P_{CO}}} \right)}^2}}}{{{P_{C{O_2}}}}} \cr
& = \frac{{{{\left( {2'0.3} \right)}^2}}}{{\left( {0.5 - 0.3} \right)}} \cr
& = \frac{{{{\left( {0.6} \right)}^2}}}{{\left( {0.2} \right)}} \cr
& = 1.8\,atm \cr} $$