ections 19.9 (Page), VES 19.10, 19.11. eaction: Cl(g) reaction at 25° C under nditions. Express your answer in kilojoules. AGxn-10.9 kJ Submit ✓ Correct Under standard conditions, the free energy for this reaction is negative because the equilibrium constant is greater than 1. A negative free energy also suggests that the reaction is spontaneous in the forward direction. Part B Previous Answers at equilibrium Express your answer in kilojoules. AGrxn = 0 kJ Submit Part C Previous Answers Correct Under equilibrium conditions, the value of RT InQ is always equal in magnitude but opposite in sign to the value of AGxn. Therefore, the value of A Grxn is zero and the reaction is not spontaneous in either direction, as expected for a reaction at equilibrium. PICI 2.65 atm; P₁₂ = 0.319 atm; Pcl₂ = 0.219 atm. Express your answer using one significant figure.

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This image contains a problem from a chemistry assignment focusing on Gibbs free energy and chemical reactions. 

### Instructions and Content:

1. **Part A**
   - **Task:** Calculate the standard Gibbs free energy change (ΔG°_rxn) for the reaction.
   - **Answer Given:** ΔG°_rxn = -10.9 kJ
   - **Feedback:** 
     - Correct response.
     - Explanation: Under standard conditions, the free energy for this reaction is negative because the equilibrium constant is greater than 1. A negative free energy indicates that the reaction is spontaneous in the forward direction.

2. **Part B**
   - **Task:** Determine the Gibbs free energy change (ΔG_rxn) at equilibrium.
   - **Answer Given:** ΔG_rxn = 0 kJ
   - **Feedback:**
     - Correct response.
     - Explanation: At equilibrium, the value of RT lnQ is equal in magnitude but opposite in sign to ΔG°_rxn. Hence, ΔG_rxn is zero, indicating no spontaneity in either direction for a reaction at equilibrium.

3. **Part C**
   - **Task:** Use the given pressures to analyze the reaction.
   - **Data Provided:**
     - P_Cl = 2.65 atm
     - P_1 = 0.319 atm
     - P_Ck = 0.219 atm
   - **Instructions:** Express the answer using one significant figure.

The image is an exercise in understanding chemical equilibrium and the conditions that affect the spontaneity of reactions based on Gibbs free energy.
Transcribed Image Text:This image contains a problem from a chemistry assignment focusing on Gibbs free energy and chemical reactions. ### Instructions and Content: 1. **Part A** - **Task:** Calculate the standard Gibbs free energy change (ΔG°_rxn) for the reaction. - **Answer Given:** ΔG°_rxn = -10.9 kJ - **Feedback:** - Correct response. - Explanation: Under standard conditions, the free energy for this reaction is negative because the equilibrium constant is greater than 1. A negative free energy indicates that the reaction is spontaneous in the forward direction. 2. **Part B** - **Task:** Determine the Gibbs free energy change (ΔG_rxn) at equilibrium. - **Answer Given:** ΔG_rxn = 0 kJ - **Feedback:** - Correct response. - Explanation: At equilibrium, the value of RT lnQ is equal in magnitude but opposite in sign to ΔG°_rxn. Hence, ΔG_rxn is zero, indicating no spontaneity in either direction for a reaction at equilibrium. 3. **Part C** - **Task:** Use the given pressures to analyze the reaction. - **Data Provided:** - P_Cl = 2.65 atm - P_1 = 0.319 atm - P_Ck = 0.219 atm - **Instructions:** Express the answer using one significant figure. The image is an exercise in understanding chemical equilibrium and the conditions that affect the spontaneity of reactions based on Gibbs free energy.
**Missed This?**  
Read Sections 19.9 (Page), 19.10 (Page); Watch IWEs 19.10, 19.11.

Consider the following reaction:  
\[ I_2(g) + Cl_2(g) \rightleftharpoons 2ICl(g) \]  
\[ K_p = 81.9 \text{ at } 25^\circ \text{C} \]

Calculate \(\Delta G_{rxn}\) for the reaction at 25°C under each of the following conditions.

---

### Explanation

This text is instructional and focuses on calculating the Gibbs free energy change (\(\Delta G_{rxn}\)) of a particular chemical reaction. The reaction involves iodine gas (\(I_2\)), chlorine gas (\(Cl_2\)), and iodine monochloride gas (\(ICl\)) in equilibrium. The equilibrium constant (\(K_p\)) is given as 81.9 at a temperature of 25°C.

Students are encouraged to review specific sections and interactive web elements (IWEs) for further understanding before attempting the calculation.
Transcribed Image Text:**Missed This?** Read Sections 19.9 (Page), 19.10 (Page); Watch IWEs 19.10, 19.11. Consider the following reaction: \[ I_2(g) + Cl_2(g) \rightleftharpoons 2ICl(g) \] \[ K_p = 81.9 \text{ at } 25^\circ \text{C} \] Calculate \(\Delta G_{rxn}\) for the reaction at 25°C under each of the following conditions. --- ### Explanation This text is instructional and focuses on calculating the Gibbs free energy change (\(\Delta G_{rxn}\)) of a particular chemical reaction. The reaction involves iodine gas (\(I_2\)), chlorine gas (\(Cl_2\)), and iodine monochloride gas (\(ICl\)) in equilibrium. The equilibrium constant (\(K_p\)) is given as 81.9 at a temperature of 25°C. Students are encouraged to review specific sections and interactive web elements (IWEs) for further understanding before attempting the calculation.
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