Question

Solids $CaC{O}_3$ and CaO and gaseous $C{O}_2$ are placed in a vessel and allowed to reach equilibrium

$CaO\left(s\right)+C{O}_2\left(g\right)\rightleftharpoons CaC{O}_3\left(s\right)\Delta H=-180kJmo{l}^{-1}$

The quantity of CaO in the vessel could be increased by:

• A. Adding more of $CaC{O}_3$
• B. Removing some of $C{O}_2$
• C. Lowering the temperature
• D. Reducing the volume of the vessel

Answer 1

The correct option is B

Removing some of $C{O}_2$

Since $K_p=1/P_{CO2}$, adding $CaC{O}_3\left(s\right)$ does not affect equilibrium. Since reaction is exothermic, lowering temperature will shift the reaction to right side. Hence, the quantity of CaO (s) will be lowered. Removing $C{O}_2\left(g\right)$ will favor equilibrium to left side. Hence, the quantity of CaO (s) will be increased. Reducing volume will increase pressure. To maintain $K_{p}or{P}_{CO2}$ to a constant value, the equilibrium will shift to right side.