Question

The value of $K _{ p }$ for the reaction, $CO _{2}( g )+ C ( s ) \rightleftharpoons 2 CO ( g )$ is $3.0$ at $1000 K$. If initially $P _{ CO _{2}}=0.48$ bar and $P _{ CO }=0$ bar and pure graphite is present, calculate the equilibrium partial pressure of $CO$ and $CO _{2}$.

• A. $0.15$ bar
• B. $ 0.33$ bar
• C. $ 0.1475$ bar
• D. $ 5.3$ bar

Answer 1

Answer: (C)

The given reaction is :-
$ \begin{array}{l} \quad CO _{2}( g )+ C ( s ) \rightleftharpoons 2 CO ( g ) \quad ; \quad K _{ P }=3 \\ \text { Initial pressure }: 0.48 \quad 0 \quad 0 \quad \text { (let) } \\ \text { At eqm : }(0.48- p ) \\ \text { Now, } K _{ P }=\frac{[ CO ]^{2}}{\left[ CO _{2}\right]} \quad K _{ P }=\frac{( pCO )^{2}}{\left( pCO _{2}\right)} \\ \Rightarrow \quad K _{ P }=\frac{(2 p )^{2}}{(0.48- p )} \\ \Rightarrow \quad 3(0.48- p )=4 p ^{2} \\ \Rightarrow \quad 4 p ^{2}+3 p -1.44=0 \\ \Rightarrow \quad p =\frac{-3 \pm \sqrt{9+23.04}}{8}=\frac{-3 \pm 5.66}{8} \\ \Rightarrow \quad p =\frac{2.66}{8}=0.3325 bar \end{array} $
So, Partial pressure of $CO$ at eqm $=2 \times 0.3325=0.665 bar$ Partial pressure of $CO _{2}$ at eqm $=0.48-0.3325=0.1475$ bar.