`Delta S` for the reaction ,
`MgCO_(3)(s)rarr MgO(s)+CO_(2)(g)` will be :

Options

A. Zero
B. `-ve`
C. `+ve`
D. `infty`

Answer

To determine the change in entropy (ΔS) for the reaction:

\[ \text{MgCO}_3(s) \rightarrow \text{MgO}(s) + \text{CO}_2(g) \]

we will follow these steps:

### Step 1: Identify the phases of the reactants and products
- The reactant is magnesium carbonate (MgCO₃), which is a solid (s).
- The products are magnesium oxide (MgO), which is also a solid (s), and carbon dioxide (CO₂), which is a gas (g).

### Step 2: Analyze the change in the number of gas moles
- In the reactants, there are **0 moles of gas** (since MgCO₃ is a solid).
- In the products, there is **1 mole of gas** (from CO₂).

### Step 3: Apply the concept of entropy
- Entropy (S) is a measure of disorder or randomness in a system. Gases have higher entropy than solids due to their greater freedom of movement and higher randomness.
- Since the reaction produces 1 mole of gas from solids, we can conclude that the entropy of the system increases.

### Step 4: Calculate the change in entropy (ΔS)
- The change in entropy can be calculated using the formula:

\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \]

- Since we are considering only the gaseous products and reactants, we can simplify this to:

\[ \Delta S = \text{Number of moles of gas in products} - \text{Number of moles of gas in reactants} \]

- Therefore:

\[ \Delta S = 1 - 0 = 1 \]

### Conclusion:
The change in entropy (ΔS) for the reaction is positive, indicating an increase in disorder due to the formation of a gas from solids.

### Final Answer:
\[ \Delta S > 0 \quad \text{(Positive change in entropy)} \]

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