There are two crystalline forms of PbO , one is yellow and the other isred. The standard enthalpies of formation of these twoforms are - 217.3 and -219.0 kJ per mole respectively. Calculate the enthalpy change for the solid - solid phase transition.
PbO( yellow ) `rarr ` PbO ( red)

Explanation

To calculate the enthalpy change for the solid-solid phase transition of PbO from yellow to red, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Standard Enthalpies of Formation**:
- The standard enthalpy of formation for PbO (yellow) is given as \( \Delta H_f^{\text{yellow}} = -217.3 \, \text{kJ/mol} \).
- The standard enthalpy of formation for PbO (red) is given as \( \Delta H_f^{\text{red}} = -219.0 \, \text{kJ/mol} \).

2. **Write the Reaction for the Phase Transition**:
- The phase transition we are interested in is:
\[ \text{PbO (yellow)} \rightarrow \text{PbO (red)} \]

3. **Use Hess's Law**:
- According to Hess's law, the enthalpy change for the transition can be calculated by taking the difference between the enthalpies of formation of the two forms:
\[ \Delta H_{\text{transition}} = \Delta H_f^{\text{red}} - \Delta H_f^{\text{yellow}} \]

4. **Substitute the Values**:
- Substitute the values of the enthalpies:
\[ \Delta H_{\text{transition}} = (-219.0 \, \text{kJ/mol}) - (-217.3 \, \text{kJ/mol}) \]

5. **Calculate the Result**:
- Simplifying the equation:
\[ \Delta H_{\text{transition}} = -219.0 + 217.3 = -1.7 \, \text{kJ/mol} \]

6. **Final Answer**:
- The enthalpy change for the solid-solid phase transition from PbO (yellow) to PbO (red) is:
\[ \Delta H_{\text{transition}} = -1.7 \, \text{kJ/mol} \]