A contributions of both heat (enthalpy) and randomness(entropy) shall be considered to the overall spontaneity of process.When deciding about the spontaneity of a chemical reaction or other process, we define a quantity called the Gibb's energy change `(DeltaG)`.
`DeltaG=DeltaH-TDeltaS`
where,`DeltaH`=Enthalpy change, `DeltaS`=Entropy change , T=Temperature in kelvin.
If `DeltaGlt0`, Process is spontaneous , `DeltaG=0`, Process is at equilibrium , `DeltaGgt0`, Process is non-spontaneous.
Quick lime (CaO) is produced by heating limestone `(CaCO_3)` to drive off `CO_2` gas.
`CaCO_3(s)toCaO(s)+CO_2(g),DeltaH^@=180 KJ,DeltaS^@=150 J//K`
Assuming that variation of enthalpy change and entropy change with temperature to be negligible, choose the correct option :

Options

A. Decomposition of `CaCO_s (s)` is never sppotaneous
B. Decomposition of `CaCO_s (s)` becomes spontaneous when temperature is less than `900^@C`
C. Decomposition of `CaCO_3 (s)` becomes non-spontaneous when temperature is greater than `1000^@`
D. Decomposition of `CaCO_3 (s)` becomes spontaneous when temperature is greater than `927^@C`

Explanation

To determine the spontaneity of the decomposition of calcium carbonate (CaCO₃) into quick lime (CaO) and carbon dioxide (CO₂), we will use the Gibbs free energy change equation:

\[ \Delta G = \Delta H - T \Delta S \]

### Step 1: Identify Given Values
From the problem, we have:
- \(\Delta H = 180 \, \text{kJ} = 180 \times 10^3 \, \text{J}\) (since 1 kJ = 1000 J)
- \(\Delta S = 150 \, \text{J/K}\)

### Step 2: Calculate the Temperature at Which the Reaction is at Equilibrium
At equilibrium, \(\Delta G = 0\). Thus, we can set up the equation:

\[ 0 = \Delta H - T \Delta S \]

Rearranging gives:

\[ T = \frac{\Delta H}{\Delta S} \]

Substituting the values:

\[ T = \frac{180 \times 10^3 \, \text{J}}{150 \, \text{J/K}} = 1200 \, \text{K} \]

### Step 3: Convert Temperature to Celsius
To convert from Kelvin to Celsius:

\[ T(°C) = T(K) - 273 \]

Calculating:

\[ T(°C) = 1200 - 273 = 927 \, °C \]

### Step 4: Determine Conditions for Spontaneity
The reaction will be spontaneous when \(\Delta G < 0\). This occurs when:

\[ T > 1200 \, \text{K} \quad \text{or} \quad T > 927 \, °C \]

### Step 5: Analyze the Options
Now we analyze the given options:
1. Decomposition of CaCO₃ solid is never spontaneous. (Incorrect)
2. Decomposition of calcium carbonate becomes spontaneous when temperature is less than 900 °C. (Incorrect)
3. Decomposition of CaCO₃ solid becomes non-spontaneous when temperature is greater than 1000 °C. (Incorrect)
4. Decomposition of calcium carbonate becomes spontaneous when temperature is greater than 927 °C. (Correct)

### Conclusion
The correct option is:
**Decomposition of calcium carbonate becomes spontaneous when temperature is greater than 927 °C.**

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