Question

A vessel at $1000K$ contains $C{O}_2$ with a pressure of $0.5atm$. Some of $C{O}_2$ is converted into $CO$ on addition of graphite. The value of ${}^{\prime }{K}_p^{\prime }$ at equilibrium when total pressure is $0.8atm$ will be ?

• A. $0.18atm$
• B. $1.8atm$
• C. $2atm$
• D. $1atm$

Answer 1

The correct option is B $1.8atm$

The given reaction is,

$CO₂+C\to 2CO$

Initial pressure of $CO₂$ , $P_i=0.5atm$

Initial pressure of $CO=0$

Final pressure of $CO₂=(0.5-x)atm$

Final pressure of $CO=2xatm$

Total pressure at equilibrium = Partial pressures of $CO₂$ and $CO$

⇒ $P_{eq}=(0.5-x)+2x=0.8atm$

⇒ $x=\mathrm{0.3.}$

∴ Partial pressures of $CO₂=0.2atm$

and Partial pressures of $CO=0.6atm$

Thus, $K_p$ = $\frac{P_{CO}^2}{P_{C{O}_2}}$ = $\frac{(0.6{)}^2}{0.2}$ = $1.8atm$

Hence, option B is correct