Question

The $\Delta_f\text{H}^\ominus$ of $\text{BaCO}_3\text{(s)}$, BaO(s) and $\text{CO}_2\text{(g)}$ is respectively $-1216.3$, $-553.5$ and $-393.5\text{ kJ mol}^{-1}$. What is the value of $x$ (in $\text{kJ mol}^{-1}$) in the following reaction? \[ \text{BaCO}_3\text{(s)} \xrightarrow{\Delta} \text{BaO(s)}+\text{CO}_2\text{(g)}-x \]

• A. $-269.3$
• B. $269.3$
• C. $2163$
• D. $-2163$

Hint:

For decomposition reactions, a positive value of $\Delta H$ indicates that heat must be supplied to break the compound into simpler substances. Since metal carbonates are generally thermally stable, their decomposition reactions are usually endothermic.

Answer 1

The Correct Option is B

Step 1: Recall Hess's law for reaction enthalpy.
The reaction enthalpy is the sum of formation enthalpies of products minus that of reactants, \[ \Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants}) \]
Step 2: Write the decomposition.
The reaction is $\text{BaCO}_3(s) \rightarrow \text{BaO}(s) + \text{CO}_2(g)$.
Step 3: Apply the formula.
\[ \Delta_r H^\circ = \left[\Delta_f H^\circ(\text{BaO}) + \Delta_f H^\circ(\text{CO}_2)\right] - \Delta_f H^\circ(\text{BaCO}_3) \]
Step 4: Insert the given values.
\[ \Delta_r H^\circ = \left[(-553.5) + (-393.5)\right] - (-1216.3) \]
Step 5: Simplify the arithmetic.
The products sum to $-947.0$, so $\Delta_r H^\circ = -947.0 + 1216.3 = 269.3$ kJ per mol.
Step 6: Match to $x$.
The reaction absorbs this much energy, so $x = 269.3$ kJ per mol, which is option 2.
\[ \boxed{x = 269.3\ \text{kJ mol}^{-1}} \]