Question

The standard enthalpy of formation of $CO_2$(g), ${CaO(s)}$ and ${CaCO_3(s)}$ are -393, -634, -1210 kJ mol$^{-1}$ respectively. If all the substans are in standard state, the standard enthalpy of decomposition of calcium carbonate to ${CaO(s)}$ and ${CO_2(g)}$ (in kJ mol$^{-1}$) is

• A. $969$
• B. $183$
• C. $-969$
• D. $-183$

Hint:

Using Enthalpies of Formation. To calculate $\Delta H^\circ$ for any reaction: \[ \Delta H^\circ = \sum \Delta H^\circ_f(\textproducts) - \sum \Delta H^\circ_f(\textreactants) \] Make sure you use standard values and balance the reaction.

Answer 1

The Correct Option is B

We are given a decomposition reaction:

\[ \mathrm{CaCO_3 (s) \rightarrow CaO (s) + CO_2 (g)} \]

To find the standard enthalpy change (\( \Delta H^\circ \)) for this decomposition, apply Hess's law:

\[ \Delta H^\circ = \sum \Delta H^\circ_f(\text{products}) - \sum \Delta H^\circ_f(\text{reactants}) \]

Given:

\[ \begin{aligned} \Delta H^\circ_f(\mathrm{CO_2 (g)}) &= -393 \text{ kJ/mol} \\ \Delta H^\circ_f(\mathrm{CaO (s)}) &= -634 \text{ kJ/mol} \\ \Delta H^\circ_f(\mathrm{CaCO_3 (s)}) &= -1210 \text{ kJ/mol} \end{aligned} \]

Substitute into the formula:

\[ \Delta H^\circ = [(-634) + (-393)] - (-1210) = -1027 + 1210 = 183 \text{ kJ/mol} \]

Hence, the standard enthalpy of decomposition is \( 183 \text{ kJ/mol} \), indicating an endothermic reaction.

Answer 2

Step 1: Write the reaction for decomposition of calcium carbonate
\[ \mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)} \]

Step 2: Use the standard enthalpy of formation values
Given:
\(\Delta H_f^\circ (\mathrm{CaCO_3}) = -1210\, \mathrm{kJ/mol}\)
\(\Delta H_f^\circ (\mathrm{CaO}) = -634\, \mathrm{kJ/mol}\)
\(\Delta H_f^\circ (\mathrm{CO_2}) = -393\, \mathrm{kJ/mol}\)

Step 3: Apply the formula for enthalpy change of reaction
\[ \Delta H^\circ_{\text{reaction}} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}) \]
\[ = [\Delta H_f^\circ(\mathrm{CaO}) + \Delta H_f^\circ(\mathrm{CO_2})] - \Delta H_f^\circ(\mathrm{CaCO_3}) \]
\[ = (-634) + (-393) - (-1210) = (-1027) + 1210 = 183\, \mathrm{kJ/mol} \]

Step 4: Conclusion
The standard enthalpy of decomposition of calcium carbonate is \(+183\, \mathrm{kJ/mol}\). This positive value indicates the reaction is endothermic and requires heat input to proceed.