Question

Consider the following reaction: \[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \quad \Delta H = 178\text{ kJ} \] The standard enthalpy of formation of $\text{CaCO}_3(s)$ and $\text{CO}_2(g)$ is $-1207$ and $-393\text{ kJ}\cdot\text{mol}^{-1}$ respectively. What is $\Delta_f H^\ominus$ (in $\text{kJ}\cdot\text{mol}^{-1}$) of $\text{CaO}(s)$?

• A. $-636$
• B. $+636$
• C. $-814$
• D. $+814$

Hint:

Always ensure proper tracking of signs ($+$ and $-$) in thermochemistry problems.
A common error is forgetting to change the sign of the reactants when subtracting.

Answer 1

The Correct Option is A

Step 1: Set up Hess's law for the reaction.
\[ \Delta_r H^\ominus = \Delta_f H^\ominus(\text{CaO}) + \Delta_f H^\ominus(\text{CO}_2) - \Delta_f H^\ominus(\text{CaCO}_3) \]
Step 2: Substitute the known numbers.
$178 = \Delta_f H^\ominus(\text{CaO}) + (-393) - (-1207)$, which simplifies to $178 = \Delta_f H^\ominus(\text{CaO}) + 814$.
Step 3: Isolate $\Delta_f H^\ominus(\text{CaO})$.
Moving 814 to the other side gives $\Delta_f H^\ominus(\text{CaO}) = 178 - 814$.
\[ \boxed{\Delta_f H^\ominus(\text{CaO}) = -636 \text{ kJ/mol}} \]