Question

Calculate the heat produced (in $\text{kJ}$ ) when $\text{224} \, \, \text{gm}$ of $\text{CaO}$ is completely converted to $\text{CaCO}_{3}$ by reaction with $\text{CO}_{\text{2}}$ at $\text{27}^{\text{o}} \text{C}$ in a container of fixed volume.
Given : $\Delta H _{ f }^{\circ}\left[ CaCO _{3}( s )\right]=-1207\, kJ / mol ; \Delta H _{ f }^{\circ}[ CaO ( s )]=-635\, kJ / mol$
$\Delta H _{ f }^{\circ}\left( CO _{2}, g \right)=-394\, kJ / mol ; \left[\right.$ Use $\left.R =8.3 \,K ^{-1} \,mol ^{-1}\right]$

• A. $-702.04\, kJ$
• B. $721.96 \,kJ$
• C. $712\, kJ$
• D. $721\, kJ$

Answer 1

Answer: (A)

$ CaO ( s ) + CO _{2}( g ) \rightarrow CaCO _{3}( s ) $
$ \Delta_{ r } H ^{\circ} =\Delta H _{ f }^{\circ}\left( CaCO _{3}\right)-\Delta H _{ f }^{\circ}( CaO )-\Delta H _{ f }^{\circ}\left( CO _{2}\right) $
$=-1207-(-635)-(-394) $
$=-178 kJ / mol $
$ \therefore \quad \Delta E _{ r }^{\circ}=\Delta H _{ r }^{\circ}-\Delta n _{ g } RT $
$ \Delta E _{ r }^{\circ} =-178-\frac{(-1) \times 8.3 \times 300}{1000} $
$=-175.51 \,kJ $
$n _{ CaO } =\frac{224}{56}=4 $
$ \therefore q _{ v } = n \cdot \Delta r _{ r } E =4 \times(-175.51) $
$=-702.04 \,kJ$