Question

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

• A. If $T$ increases, reaction proceeds in forward direction.
• B. If $T$ increases, reaction proceeds in backward direction.
• C. If $T$ decreases, reaction proceeds in forward direction.
• D. If $T$ increases, reaction remains at equilibrium.

Answer 1

The correct option is A If $T$ increases, reaction proceeds in forward direction.

${\Delta }_r{H}^{\ominus }={\Delta }_r{H}^{\ominus }\left[CaO\right(s\left)\right]+{\Delta }_r{H}^{\ominus }\left[C{O}_2\right(9\left)\right]-{\Delta }_r{H}^{\ominus }\left[CaC{O}_3\right(s\left)\right]$
$\therefore {\Delta }_r{H}^{\ominus }=178.3kJmo{l}^{-1}$
The reaction is endothermic.
Hence, according to Le-Chatelier's principle, reaction will proceed in forward direction on increasing temperature.