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:

Whenever enthalpies of formation are given, remember the shortcut: \[ \Delta H^\circ_{\text{reaction}} = \Sigma \Delta H^\circ_f(\text{products}) - \Sigma \Delta H^\circ_f(\text{reactants}) \] A positive value of $\Delta H$ indicates an endothermic reaction, while a negative value indicates an exothermic reaction.

Answer 1

The Correct Option is B

Concept: The standard enthalpy change of a reaction is calculated using Hess's law: \[ \Delta_r H^\circ = \sum \Delta_f H^\circ(\text{Products}) - \sum \Delta_f H^\circ(\text{Reactants}) \] where $\Delta_f H^\circ$ denotes the standard enthalpy of formation. For the decomposition reaction \[ \text{BaCO}_3(s) \rightarrow \text{BaO}(s)+\text{CO}_2(g) \] the enthalpy change is obtained directly from the given enthalpies of formation.

Step 1: Write the expression for reaction enthalpy \[ \Delta_r H^\circ = \left[ \Delta_f H^\circ(\text{BaO}) + \Delta_f H^\circ(\text{CO}_2) \right] - \left[ \Delta_f H^\circ(\text{BaCO}_3) \right] \]

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

Step 3: Identify the value of $x$ The reaction is written as \[ \text{BaCO}_3(s) \xrightarrow{\Delta} \text{BaO}(s)+\text{CO}_2(g)-x \] which indicates that the decomposition requires energy. Therefore, \[ x=\Delta_r H^\circ \] \[ x=269.3\ \text{kJ mol}^{-1} \] Hence, \[ \boxed{x=269.3\ \text{kJ mol}^{-1}} \]