`K_(p)` = 1, For the equilibrium `CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g))`. The temperature of the reaction can be given as :

Options

A. T ` = (Delta S^(@))/(Delta H^(@))`
B. `T = (Delta H^(@))/(Delta S^(@))`
C. `T = (Delta G^(@))/(R)`
D. ` T = (Delta G^(@))/(Delta H^(@))`

Answer

To solve the problem regarding the equilibrium reaction \( \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \) with \( K_p = 1 \), we will use the relationship between Gibbs free energy, enthalpy, and entropy. Here are the steps to derive the temperature of the reaction:

### Step 1: Understand the relationship between Gibbs free energy and equilibrium constant
The Gibbs free energy change (\( \Delta G \)) at equilibrium is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]
At equilibrium, \( \Delta G = 0 \), which leads us to:
\[ 0 = \Delta H - T \Delta S \]

### Step 2: Relate Gibbs free energy to the equilibrium constant
The standard Gibbs free energy change (\( \Delta G^\circ \)) can also be expressed in terms of the equilibrium constant (\( K_p \)):
\[ \Delta G^\circ = -RT \ln K_p \]
Given that \( K_p = 1 \), we find:
\[ \ln K_p = \ln 1 = 0 \]
Thus,
\[ \Delta G^\circ = -RT \cdot 0 = 0 \]

### Step 3: Set up the equation for \( \Delta H \) and \( \Delta S \)
Since we have established that \( \Delta G^\circ = 0 \), we can substitute this back into the first equation:
\[ 0 = \Delta H - T \Delta S \]
Rearranging gives:
\[ \Delta H = T \Delta S \]

### Step 4: Solve for temperature \( T \)
From the equation \( \Delta H = T \Delta S \), we can isolate \( T \):
\[ T = \frac{\Delta H}{\Delta S} \]

### Conclusion
The temperature of the reaction can be expressed as:
\[ T = \frac{\Delta H}{\Delta S} \]
This means that the temperature at which the equilibrium constant \( K_p = 1 \) is determined by the ratio of the change in enthalpy to the change in entropy.

### Final Answer
The temperature of the reaction can be given as \( T = \frac{\Delta H}{\Delta S} \).

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