Consider the reaction 2NOcg)t02(g)-2NO2C9). alusing data from append ik C, Calculate AG°at 298 K b) calculate AGat 298 Kif the partial pressure ot allgases are 33.4K Pa NOCAHE 's 90.37)(5°"s 210.62)(Ge is 86.71) NO2(AH6's 33.84)(s°ig 240.45)(BG's is 51.84) O2 (AHi is 0)(S"is 205.0) (AGiEs 0)
Consider the reaction 2NOcg)t02(g)-2NO2C9). alusing data from append ik C, Calculate AG°at 298 K b) calculate AGat 298 Kif the partial pressure ot allgases are 33.4K Pa NOCAHE 's 90.37)(5°"s 210.62)(Ge is 86.71) NO2(AH6's 33.84)(s°ig 240.45)(BG's is 51.84) O2 (AHi is 0)(S"is 205.0) (AGiEs 0)
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![### Thermodynamics Problem: Calculating ΔG° at 298 K for a Chemical Reaction
**Problem Statement:**
Consider the reaction:
\[ 2 \text{NO(g)} + \text{O}_2(\text{g}) \rightarrow 2 \text{NO}_2(\text{g}) \]
a) Using data from Appendix C, calculate ΔG° at 298 K.
b) Calculate ΔG at 298 K if the partial pressure of all gases is 33.4 kPa.
**Thermodynamic Data from Appendix C:**
- **NO(g):**
- \( \Delta H^\circ_f \) = 90.37 kJ/mol
- \( S^\circ \) = 210.62 J/(mol·K)
- \( \Delta G^\circ_f \) = 86.71 kJ/mol
- **NO₂(g):**
- \( \Delta H^\circ_f \) = 33.84 kJ/mol
- \( S^\circ \) = 240.45 J/(mol·K)
- \( \Delta G^\circ_f \) = 51.84 kJ/mol
- **O₂(g):**
- \( \Delta H^\circ_f \) = 0 kJ/mol
- \( S^\circ \) = 205.0 J/(mol·K)
- \( \Delta G^\circ_f \) = 0 kJ/mol
**Instructions:**
1. Use the provided thermodynamic values to calculate the standard Gibbs free energy change (ΔG°) for the reaction at 298 K using the following formula:
\[ \Delta G^\circ = \sum (\Delta G^\circ_f \text{ of products}) - \sum (\Delta G^\circ_f \text{ of reactants}) \]
2. Adjust for non-standard conditions using the appropriate thermodynamic equations if necessary.
_Remember to account for stoichiometry and units when performing calculations._](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ef85c79-8e8f-4d9a-890a-b67fe0c4488a%2Ff7ba2611-36e7-468c-ab51-d06c2ceb0610%2F24whrci_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Thermodynamics Problem: Calculating ΔG° at 298 K for a Chemical Reaction
**Problem Statement:**
Consider the reaction:
\[ 2 \text{NO(g)} + \text{O}_2(\text{g}) \rightarrow 2 \text{NO}_2(\text{g}) \]
a) Using data from Appendix C, calculate ΔG° at 298 K.
b) Calculate ΔG at 298 K if the partial pressure of all gases is 33.4 kPa.
**Thermodynamic Data from Appendix C:**
- **NO(g):**
- \( \Delta H^\circ_f \) = 90.37 kJ/mol
- \( S^\circ \) = 210.62 J/(mol·K)
- \( \Delta G^\circ_f \) = 86.71 kJ/mol
- **NO₂(g):**
- \( \Delta H^\circ_f \) = 33.84 kJ/mol
- \( S^\circ \) = 240.45 J/(mol·K)
- \( \Delta G^\circ_f \) = 51.84 kJ/mol
- **O₂(g):**
- \( \Delta H^\circ_f \) = 0 kJ/mol
- \( S^\circ \) = 205.0 J/(mol·K)
- \( \Delta G^\circ_f \) = 0 kJ/mol
**Instructions:**
1. Use the provided thermodynamic values to calculate the standard Gibbs free energy change (ΔG°) for the reaction at 298 K using the following formula:
\[ \Delta G^\circ = \sum (\Delta G^\circ_f \text{ of products}) - \sum (\Delta G^\circ_f \text{ of reactants}) \]
2. Adjust for non-standard conditions using the appropriate thermodynamic equations if necessary.
_Remember to account for stoichiometry and units when performing calculations._
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