Question

The $\Delta_f\text{H}^\ominus$ of $\text{BaCO}_3\text{(s)}$, $\text{BaO(s)}$ 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

Concept: The standard enthalpy change of a reaction is given by \[ \Delta_rH^\circ = \sum \Delta_fH^\circ(\text{Products}) - \sum \Delta_fH^\circ(\text{Reactants}) \] For the decomposition reaction \[ \text{BaCO}_3(s) \rightarrow \text{BaO}(s)+\text{CO}_2(g) \] the reaction enthalpy is obtained from the given standard enthalpies of formation.

Step 1: Write the enthalpy expression \[ \Delta_rH^\circ = \left[ \Delta_fH^\circ(\text{BaO}) + \Delta_fH^\circ(\text{CO}_2) \right] - \left[ \Delta_fH^\circ(\text{BaCO}_3) \right] \]

Step 2: Substitute the given values \[ \Delta_rH^\circ = \left[ (-553.5)+(-393.5) \right] - (-1216.3) \] \[ \Delta_rH^\circ = (-947.0)+1216.3 \] \[ \Delta_rH^\circ = 269.3\ \text{kJ mol}^{-1} \]

Step 3: Relate the result to $x$ The decomposition reaction is endothermic. Therefore, \[ x=\Delta_rH^\circ \] \[ x=269.3\ \text{kJ mol}^{-1} \] Hence, \[ \boxed{x=269.3\ \text{kJ mol}^{-1}} \]