Question

calculate $\delta g^\circ$ for the reaction \\( \text{cao (s)} + \text{co}_2 \text{(g)} \
ightarrow \text{caco}_3 \text{(s)} \\)\
\

substance	$\delta g^\circ$f (kj/mol)	\
-----------	---------------------------	\
cao (s)	-603.3	\
co₂ (g)	-394.4	\
caco₃ (s)	-1128.2

Explanation:

Step1: Recall the formula for $\Delta G^\circ$ of a reaction

The formula for the standard Gibbs free energy change ($\Delta G^\circ$) of a reaction is $\Delta G^\circ=\sum n\Delta G^\circ_f(\text{products})-\sum m\Delta G^\circ_f(\text{reactants})$, where $n$ and $m$ are the stoichiometric coefficients of products and reactants respectively. For the reaction $\ce{CaO (s) + CO2 (g) -> CaCO3 (s)}$, the stoichiometric coefficients of all substances are 1. So the formula simplifies to $\Delta G^\circ=\Delta G^\circ_f(\ce{CaCO3 (s)})-[\Delta G^\circ_f(\ce{CaO (s)})+\Delta G^\circ_f(\ce{CO2 (g)})]$.

Step2: Substitute the given values

We know that $\Delta G^\circ_f(\ce{CaCO3 (s)})=- 1128.2\space kJ/mol$, $\Delta G^\circ_f(\ce{CaO (s)})=-603.3\space kJ/mol$ and $\Delta G^\circ_f(\ce{CO2 (g)})=-394.4\space kJ/mol$. Substituting these values into the formula:

$\Delta G^\circ=-1128.2-(-603.3 - 394.4)$

First, calculate the sum of the reactants' $\Delta G^\circ_f$: $-603.3+(-394.4)=-603.3 - 394.4=-997.7\space kJ/mol$

Then, $\Delta G^\circ=-1128.2-(-997.7)=-1128.2 + 997.7=-130.5\space kJ/mol$

Answer:

$\boldsymbol{-130.5\space kJ/mol}$