`S_(("rhombic"))+O_(2(g))rarr SO_(2(g)) , Delta H =-297.5 K J`
`S_(("monoclinic"))+O_(2(g))rarr SO_(2) , Delta H = - 300 K J`
The data can predict that -

Options

A. Rhombic sulphur is yellow in colour
B. Monoclinic sulphur has metallic lustre
C. Monoclinic sulphur is more stable
D. `Delta H_(Transition) of S(R) to S(M)` is endothermic process

Answer

To solve the problem, we need to analyze the given reactions and their respective enthalpy changes. The reactions are as follows:

1. **Rhombic Sulfur Reaction:**
\[ S_{\text{(rhombic)}} + O_2(g) \rightarrow SO_2(g), \quad \Delta H_1 = -297.5 \, \text{kJ} \]

2. **Monoclinic Sulfur Reaction:**
\[ S_{\text{(monoclinic)}} + O_2(g) \rightarrow SO_2(g), \quad \Delta H_2 = -300 \, \text{kJ} \]

### Step 1: Identify the Enthalpy Change for the Transition
To find the enthalpy change for the transition from rhombic sulfur to monoclinic sulfur, we need to calculate the difference in enthalpy changes of the two reactions. The enthalpy change for the transition (\(\Delta H_{\text{transition}}\)) can be expressed as:

\[ \Delta H_{\text{transition}} = \Delta H_1 - \Delta H_2 \]

### Step 2: Substitute the Values
Now, we substitute the values of \(\Delta H_1\) and \(\Delta H_2\):

\[ \Delta H_{\text{transition}} = (-297.5 \, \text{kJ}) - (-300 \, \text{kJ}) \]

### Step 3: Simplify the Expression
Simplifying the expression gives:

\[ \Delta H_{\text{transition}} = -297.5 + 300 \, \text{kJ} = 2.5 \, \text{kJ} \]

### Step 4: Interpret the Result
The positive value of \(\Delta H_{\text{transition}} = 2.5 \, \text{kJ}\) indicates that the transition from rhombic sulfur to monoclinic sulfur is an endothermic process. This means that heat is absorbed during the conversion of rhombic sulfur to monoclinic sulfur.

### Conclusion
Based on the analysis, the correct statement is that the transition from rhombic sulfur to monoclinic sulfur is endothermic.

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