To calculate the Gibbs free energy change [tex]\(\Delta G\)[/tex] for the given reaction at 25°C under the specified conditions, we follow a systematic approach by breaking down the steps in the process:
1. Write the given reaction and specify the partial pressures:
[tex]\[
\text{CO}_2(g) + \text{CCl}_4(g) \rightleftharpoons 2 \text{COCl}_2(g)
\][/tex]
- [tex]\( P_{\text{CO}_2} = 0.140 \, \text{atm} \)[/tex]
- [tex]\( P_{\text{CCl}_4} = 0.185 \, \text{atm} \)[/tex]
- [tex]\( P_{\text{COCl}_2} = 0.735 \, \text{atm} \)[/tex]
2. Identify the standard Gibbs free energy of formation ([tex]\(\Delta G_f^\circ\)[/tex]) for each substance:
- [tex]\(\Delta G_f^\circ (\text{CO}_2) = -394.4 \, \text{kJ/mol} \)[/tex]
- [tex]\(\Delta G_f^\circ (\text{CCl}_4) = -62.3 \, \text{kJ/mol} \)[/tex]
- [tex]\(\Delta G_f^\circ (\text{COCl}_2) = -204.9 \, \text{kJ/mol} \)[/tex]
3. Calculate the standard Gibbs free energy change ([tex]\(\Delta G^\circ\)[/tex]) of the reaction:
[tex]\[
\Delta G^\circ = \left( 2 \times \Delta G_f^\circ (\text{COCl}_2) \right) - \left( \Delta G_f^\circ (\text{CO}_2) + \Delta G_f^\circ (\text{CCl}_4) \right)
\][/tex]
Substituting the values:
[tex]\[
\Delta G^\circ = \left( 2 \times -204.9 \, \text{kJ/mol} \right) - \left( -394.4 \, \text{kJ/mol} + (-62.3 \, \text{kJ/mol}) \right)
= -409.8 \, \text{kJ/mol} - (-456.7 \, \text{kJ/mol})
= 46.9 \, \text{kJ/mol}
\][/tex]
4. Calculate the reaction quotient (Q) at the given conditions:
The reaction quotient [tex]\( Q \)[/tex] is given by:
[tex]\[
Q = \frac{(P_{\text{COCl}_2})^2}{P_{\text{CO}_2} \cdot P_{\text{CCl}_4}}
\][/tex]
Substituting the partial pressures:
[tex]\[
Q = \frac{(0.735)^2}{0.140 \times 0.185} = 20.8581
\][/tex]
5. Calculate the Gibbs free energy change ([tex]\(\Delta G\)[/tex]) at the given conditions:
[tex]\[
\Delta G = \Delta G^\circ + RT \ln Q
\][/tex]
Where:
- [tex]\( R \)[/tex] (Universal gas constant) = 0.008314 kJ/(mol K)
- [tex]\( T \)[/tex] (temperature in Kelvin) = [tex]\( 25^\circ \text{C} + 273.15 = 298.15 \, \text{K} \)[/tex]
- [tex]\( \ln Q \)[/tex] is the natural logarithm of Q
Substituting in the values:
[tex]\[
\Delta G = 46.9 \, \text{kJ/mol} + (0.008314 \, \text{kJ/(mol K)} \times 298.15 \, \text{K} \times \ln 20.8581)
\][/tex]
[tex]\[
\Delta G = 46.9 \, \text{kJ/mol} + (0.008314 \times 298.15 \times 3.036) = 54.43 \, \text{kJ/mol}
\][/tex]
Therefore, the Gibbs free energy change [tex]\(\Delta G\)[/tex] for the reaction at 25°C under the given conditions is 54.4 kJ/mol (rounded to three significant figures).
