Question

Following data is given for the reaction: $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$
${\Delta }_f{H}^o\left[CaO\right(s\left)\right]=-635.1{\text{kJ mol}}^{-1}$
${\Delta }_f{H}^o\left[C{O}_2\right(g\left)\right]=-393.5{\text{kJ mol}}^{-1}$
${\Delta }_f{H}^o\left[CaC{O}_3\right(s\left)\right]=-1206.9{\text{kJ mol}}^{-1}$
Predict the effect of temperature on the equilibrium constant of the above reaction.

• A. On increasing temperature, equilibrium will shift in forward direction
• B. On increasing temperature, equilibrium will shift in backward direction
• C. Equilibrium will remain unchanged on decreasing the temperature
• D. Equilibrium will remain unchanged on increasing the temperature

Answer 1

The correct option is A On increasing temperature, equilibrium will shift in forward direction

${\Delta }_f{H}^o={\Delta }_f{H}^o\left[CaO\right(s\left)\right]+{\Delta }_f{H}^o\left[C{O}_2\right(g\left)\right]-{\Delta }_f{H}^o\left[CaC{O}_3\right(s\left)\right]$
$\therefore {\Delta }_r{H}^o=178.3{\text{kJ mol}}^{-1}$
The reaction is endothermic. Hence, according to Le Chatlier's principle, reaction will proceed in forward direction on increasing temperature.