To determine at what temperature the reaction [tex]\( \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \)[/tex] is feasible, we need to analyze the thermodynamic properties of the reaction. We will use the Gibbs free energy equation [tex]\( \Delta G = \Delta H - T \Delta S \)[/tex]. For the reaction to be feasible, [tex]\( \Delta G \)[/tex] must be zero, which gives us the condition:
[tex]\[ T = \frac{\Delta H}{\Delta S} \][/tex]
Here's the step-by-step solution:
### Step 1: Calculate the Change in Enthalpy ([tex]\(\Delta H\)[/tex])
The change in enthalpy for the reaction is given by:
[tex]\[
\Delta H_{\text{reaction}} = \left(\Delta H_{\text{CaO}} + \Delta H_{\text{CO}_2} \right) - \Delta H_{\text{CaCO}_3}
\][/tex]
Given data in kJ/mol:
[tex]\[
\begin{array}{l}
\Delta H_{\text{CaCO}_3} = -1207 \, \text{kJ/mol} \\
\Delta H_{\text{CaO}} = -635 \, \text{kJ/mol} \\
\Delta H_{\text{CO}_2} = -394 \, \text{kJ/mol}
\end{array}
\][/tex]
Plugging in the values:
[tex]\[
\Delta H_{\text{reaction}} = \left( -635 + (-394) \right) - (-1207)
\][/tex]
[tex]\[
= -1029 + 1207
\][/tex]
[tex]\[
= 178 \, \text{kJ/mol}
\][/tex]
### Step 2: Calculate the Change in Entropy ([tex]\(\Delta S\)[/tex])
The change in entropy for the reaction is given by:
[tex]\[
\Delta S_{\text{reaction}} = \left(\Delta S_{\text{CaO}} + \Delta S_{\text{CO}_2} \right) - \Delta S_{\text{CaCO}_3}
\][/tex]
Given data in J/K mol:
[tex]\[
\begin{array}{l}
\Delta S_{\text{CaCO}_3} = 93 \, \text{J/K mol} \\
\Delta S_{\text{CaO}} = 40 \, \text{J/K mol} \\
\Delta S_{\text{CO}_2} = 214 \, \text{J/K mol}
\end{array}
\][/tex]
Plugging in the values:
[tex]\[
\Delta S_{\text{reaction}} = \left( 40 + 214 \right) - 93
\][/tex]
[tex]\[
= 254 - 93
\][/tex]
[tex]\[
= 161 \, \text{J/K mol}
\][/tex]
Convert [tex]\(\Delta S_{\text{reaction}}\)[/tex] to kJ/K mol by dividing by 1000:
[tex]\[
\Delta S_{\text{reaction}} = \frac{161}{1000} = 0.161 \, \text{kJ/K mol}
\][/tex]
### Step 3: Calculate the Temperature (T) at which [tex]\(\Delta G = 0\)[/tex]
[tex]\[
T = \frac{\Delta H_{\text{reaction}}}{\Delta S_{\text{reaction}}}
\][/tex]
Plugging in the values:
[tex]\[
T = \frac{178}{0.161}
\][/tex]
[tex]\[
T \approx 1105.6 \, \text{K}
\][/tex]
Thus, the temperature at which the reaction [tex]\( \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \)[/tex] is feasible is:
[tex]\[ \boxed{1105.6 \, \text{K}} \][/tex]
So the correct answer is:
D. 1105.6 K
