Question

The value of AH° for the following reaction is 177.8 kJ. The value of AH; for CaO(s) is
CaCO (s) → Cao (s) + COz (9)

Answer 1

To find the value of \( \Delta H_f^\circ \) (standard enthalpy of formation) for CaO(s) from the given reaction:

\[
\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)
\]

we can use the provided \( \Delta H^\circ \) of the reaction, which is 177.8 kJ.

From the reaction, we understand:

- The formation of CaCO3(s) from its elements would have an associated \( \Delta H_f^\circ \).
- The formation of CaO(s) and CO2(g) from their elements also has its own \( \Delta H_f^\circ \).

Using Hess's law, we can relate these enthalpy changes. The enthalpy change for the formation of a compound is given as:

\[
\Delta H_f^\circ \text{(CaCO}_3\text{)} = \Delta H_f^\circ \text{(CaO)} + \Delta H_f^\circ \text{(CO}_2\text{)} + \Delta H^\circ \text{(reaction)}
\]

If we isolate \( \Delta H_f^\circ \text{(CaO)} \):
\[
\Delta H_f^\circ \text{(CaO)} = \Delta H_f^\circ \text{(CaCO}_3\text{)} - \Delta H_f^\circ \text{(CO}_2\text{)} - \Delta H^\circ \text{(reaction)}
\]

1. Use known standard enthalpy of formation values:
- \( \Delta H_f^\circ \text{(CO}_2(g)) = -393.5 \) kJ/mol (standard value)
- \( \Delta H_f^\circ \text{(CaCO}_3(s)) = -1207 \) kJ/mol (standard value)

2. Plugging in the known values:
\[
\Delta H_f^\circ \text{(CaO)} = (-1207) - (-393.5) - 177.8
\]

3. Simplifying,
\[
\Delta H_f^\circ \text{(CaO)} = -1207 + 393.5 - 177.8
\]

\[
\Delta H_f^\circ \text{(CaO)} = -1207 + 215.7
\]

\[
\Delta H_f^\circ \text{(CaO)} = -991.3 \text{ kJ/mol}
\]

Thus, the standard enthalpy of formation for \( \text{CaO(s)} \) is approximately \( -991.3 \) kJ/mol.