At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
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).
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).
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
