Find the best solutions to your problems with the help of IDNLearn.com's expert users. Our experts are ready to provide prompt and detailed answers to any questions you may have.
Sagot :
To calculate the standard free energy change ([tex]\(\Delta G^\ominus\)[/tex]) of the given reactions, we'll use the standard free energy of formation ([tex]\(\Delta G_f^\ominus\)[/tex]) values provided.
The standard free energy change for a reaction can be calculated using the following equation:
[tex]\[ \Delta G^\ominus = \sum \Delta G_f^\ominus \text{(products)} - \sum \Delta G_f^\ominus \text{(reactants)} \][/tex]
Let's apply this formula to each reaction.
### Reaction i
Reaction:
[tex]\[ SiO_2 (s) + 4 HF (g) \rightarrow SiF_4 (g) + 2 H_2O (l) \][/tex]
Given [tex]\(\Delta G_f^\ominus\)[/tex] values:
- [tex]\(\Delta G_f^\ominus [SiO_2 (s)] = -856.7 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [HF (g)] = -273.2 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [SiF_4 (g)] = -1572.7 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [H_2O (l)] = -69.9 \text{ kJ/mol}\)[/tex]
Calculation:
[tex]\[ \Delta G^\ominus(\text{i}) = \left[ \Delta G_f^\ominus(\text{SiF}_4) + (2) \Delta G_f^\ominus(\text{H}_2\text{O}) \right] - \left[ \Delta G_f^\ominus(\text{SiO}_2) + (4) \Delta G_f^\ominus(\text{HF}) \right] \][/tex]
Plugging in the given values:
[tex]\[ \Delta G^\ominus(\text{i}) = \left[ -1572.7 + 2(-69.9) \right] - \left[ -856.7 + 4(-273.2) \right] \][/tex]
[tex]\[ \Delta G^\ominus(\text{i}) = \left[ -1572.7 - 139.8 \right] - \left[ -856.7 - 1092.8 \right] \][/tex]
[tex]\[ \Delta G^\ominus(\text{i}) = -1712.5 - (-1949.5) \][/tex]
[tex]\[ \Delta G^\ominus(\text{i}) = -1712.5 + 1949.5 = 237 \text{ kJ/mol} \][/tex]
### Reaction ii
Reaction:
[tex]\[ SrCO_3 (s) \rightarrow SrO (s) + CO_2 (g) \][/tex]
Given [tex]\(\Delta G_f^\ominus\)[/tex] values:
- [tex]\(\Delta G_f^\ominus [SrCO_3 (s)] = -1140.4 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [SrO (s)] = -561.9 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [CO_2 (g)] = -394.4 \text{ kJ/mol}\)[/tex]
Calculation:
[tex]\[ \Delta G^\ominus(\text{ii}) = \left[ \Delta G_f^\ominus(\text{SrO}) + \Delta G_f^\ominus(\text{CO}_2) \right] - \left[ \Delta G_f^\ominus(\text{SrCO}_3) \right] \][/tex]
Plugging in the given values:
[tex]\[ \Delta G^\ominus(\text{ii}) = \left[ -561.9 + (-394.4) \right] - \left[ -1140.4 \right] \][/tex]
[tex]\[ \Delta G^\ominus(\text{ii}) = (-561.9 - 394.4) - (-1140.4) \][/tex]
[tex]\[ \Delta G^\ominus(\text{ii}) = -956.3 - (-1140.4) \][/tex]
[tex]\[ \Delta G^\ominus(\text{ii}) = -956.3 + 1140.4 = 184.1 \text{ kJ/mol} \][/tex]
Hence, the standard free energy changes for the reactions are:
- For reaction i: [tex]\(\Delta G^\ominus = 237 \text{ kJ/mol}\)[/tex]
- For reaction ii: [tex]\(\Delta G^\ominus = 184.1 \text{ kJ/mol}\)[/tex]
The standard free energy change for a reaction can be calculated using the following equation:
[tex]\[ \Delta G^\ominus = \sum \Delta G_f^\ominus \text{(products)} - \sum \Delta G_f^\ominus \text{(reactants)} \][/tex]
Let's apply this formula to each reaction.
### Reaction i
Reaction:
[tex]\[ SiO_2 (s) + 4 HF (g) \rightarrow SiF_4 (g) + 2 H_2O (l) \][/tex]
Given [tex]\(\Delta G_f^\ominus\)[/tex] values:
- [tex]\(\Delta G_f^\ominus [SiO_2 (s)] = -856.7 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [HF (g)] = -273.2 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [SiF_4 (g)] = -1572.7 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [H_2O (l)] = -69.9 \text{ kJ/mol}\)[/tex]
Calculation:
[tex]\[ \Delta G^\ominus(\text{i}) = \left[ \Delta G_f^\ominus(\text{SiF}_4) + (2) \Delta G_f^\ominus(\text{H}_2\text{O}) \right] - \left[ \Delta G_f^\ominus(\text{SiO}_2) + (4) \Delta G_f^\ominus(\text{HF}) \right] \][/tex]
Plugging in the given values:
[tex]\[ \Delta G^\ominus(\text{i}) = \left[ -1572.7 + 2(-69.9) \right] - \left[ -856.7 + 4(-273.2) \right] \][/tex]
[tex]\[ \Delta G^\ominus(\text{i}) = \left[ -1572.7 - 139.8 \right] - \left[ -856.7 - 1092.8 \right] \][/tex]
[tex]\[ \Delta G^\ominus(\text{i}) = -1712.5 - (-1949.5) \][/tex]
[tex]\[ \Delta G^\ominus(\text{i}) = -1712.5 + 1949.5 = 237 \text{ kJ/mol} \][/tex]
### Reaction ii
Reaction:
[tex]\[ SrCO_3 (s) \rightarrow SrO (s) + CO_2 (g) \][/tex]
Given [tex]\(\Delta G_f^\ominus\)[/tex] values:
- [tex]\(\Delta G_f^\ominus [SrCO_3 (s)] = -1140.4 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [SrO (s)] = -561.9 \text{ kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\ominus [CO_2 (g)] = -394.4 \text{ kJ/mol}\)[/tex]
Calculation:
[tex]\[ \Delta G^\ominus(\text{ii}) = \left[ \Delta G_f^\ominus(\text{SrO}) + \Delta G_f^\ominus(\text{CO}_2) \right] - \left[ \Delta G_f^\ominus(\text{SrCO}_3) \right] \][/tex]
Plugging in the given values:
[tex]\[ \Delta G^\ominus(\text{ii}) = \left[ -561.9 + (-394.4) \right] - \left[ -1140.4 \right] \][/tex]
[tex]\[ \Delta G^\ominus(\text{ii}) = (-561.9 - 394.4) - (-1140.4) \][/tex]
[tex]\[ \Delta G^\ominus(\text{ii}) = -956.3 - (-1140.4) \][/tex]
[tex]\[ \Delta G^\ominus(\text{ii}) = -956.3 + 1140.4 = 184.1 \text{ kJ/mol} \][/tex]
Hence, the standard free energy changes for the reactions are:
- For reaction i: [tex]\(\Delta G^\ominus = 237 \text{ kJ/mol}\)[/tex]
- For reaction ii: [tex]\(\Delta G^\ominus = 184.1 \text{ kJ/mol}\)[/tex]
We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Your questions find answers at IDNLearn.com. Thanks for visiting, and come back for more accurate and reliable solutions.