If for the heterogeneous equilibrium`CaCO_3(s) hArr CaO(s) +CO_2(g)` , the equilibrium constant `K_c=1`, at 1 atm what would be the temperature of the reaction?

Options

A. `T=(DeltaS^@)/(DeltaH^@)`
B. `T=(DeltaH^@)/(DeltaS^@)`
C. `T=(DeltaH^@)/(DeltaG^@)`
D. `T=(DeltaG^@)/(DeltaH^@)`

Explanation

To solve the problem regarding the equilibrium of the reaction:

\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]

with an equilibrium constant \( K_c = 1 \) at 1 atm, we can follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_c \).

For the reaction, the equilibrium constant \( K_c \) is defined only in terms of the gaseous products since solids do not appear in the expression. Therefore, we have:

\[ K_c = [\text{CO}_2] \]

### Step 2: Substitute the given value of \( K_c \).

We are given that \( K_c = 1 \). Thus, we can write:

\[ [\text{CO}_2] = 1 \text{ atm} \]

### Step 3: Relate \( K_c \) to Gibbs free energy change \( \Delta G^\circ \).

The relationship between the standard Gibbs free energy change \( \Delta G^\circ \) and the equilibrium constant \( K_c \) is given by the equation:

\[ \Delta G^\circ = -RT \ln K_c \]

### Step 4: Substitute \( K_c = 1 \) into the Gibbs free energy equation.

Since \( K_c = 1 \), we have:

\[ \Delta G^\circ = -RT \ln(1) \]

Knowing that \( \ln(1) = 0 \), we find:

\[ \Delta G^\circ = 0 \]

### Step 5: Use the relationship between \( \Delta G^\circ \), \( \Delta H^\circ \), and \( \Delta S^\circ \).

We can also express \( \Delta G^\circ \) in terms of enthalpy and entropy changes:

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

### Step 6: Set \( \Delta G^\circ = 0 \) and solve for temperature \( T \).

Since we have established that \( \Delta G^\circ = 0 \):

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

Rearranging gives:

\[ T \Delta S^\circ = \Delta H^\circ \]

Thus, we can express the temperature as:

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

### Conclusion:

Since we do not have specific values for \( \Delta H^\circ \) and \( \Delta S^\circ \), we conclude that the temperature at which the reaction has \( K_c = 1 \) can be expressed as:

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

This indicates that the temperature depends on the specific enthalpy and entropy changes of the reaction.