Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
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).
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.