In conversion of limestone to lime , `CaCO_3(s) rarrCaO(s)+CO_2(g)` the values of `DeltaH^@ and DeltaS^@` are +179.1 KJ/mol , 160.2J/K respectively at 298 K and 1 bar. Assuming that `DeltaH^@ and DeltaS^@` do not change with temperature , temperature above which conversion of limestone to lime will be spontaneous is

Options

A. 1008 K
B. 1200 K
C. 845 K
D. 1118 K

Answer

To determine the temperature above which the conversion of limestone to lime becomes spontaneous, we can use the Gibbs free energy equation:

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

Where:
- \(\Delta G\) is the change in Gibbs free energy,
- \(\Delta H\) is the change in enthalpy,
- \(T\) is the temperature in Kelvin,
- \(\Delta S\) is the change in entropy.

For the process to be spontaneous, \(\Delta G\) must be less than 0:

\[ \Delta G < 0 \implies \Delta H - T \Delta S < 0 \]

Rearranging this gives:

\[ T \Delta S > \Delta H \]

Thus, we can express the critical temperature \(T\) as:

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

### Step 1: Convert \(\Delta H\) to the same units as \(\Delta S\)

Given:
- \(\Delta H = +179.1 \, \text{kJ/mol}\)
- \(\Delta S = +160.2 \, \text{J/K}\)

First, we need to convert \(\Delta H\) from kJ to J:

\[ \Delta H = 179.1 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 179100 \, \text{J/mol} \]

### Step 2: Substitute values into the equation

Now we can substitute \(\Delta H\) and \(\Delta S\) into the equation for \(T\):

\[ T > \frac{179100 \, \text{J/mol}}{160.2 \, \text{J/K}} \]

### Step 3: Calculate the temperature

Now, perform the division:

\[ T > \frac{179100}{160.2} \approx 1118.4 \, \text{K} \]

### Conclusion

The temperature above which the conversion of limestone to lime will be spontaneous is approximately:

\[ T > 1118.4 \, \text{K} \]