A student investigates the reaction between Ag(s) and HNO3(aq) represented by the equation above.

Ag(s) + 4 HNO3(aq) → 3 AgNO3(aq) + NO(g) + 2 H2O(l) A student investigates the reaction between Ag(s) and HNO3(aq) represented by the equation above. Predict the sign of the entropy change, ΔS, for the reaction. Justify your answer. b. Use the information in the table below to calculate the value ΔHrxn, the standard enthalpy change for the reaction, in kJ/molrxn: Substance ΔHf° (kJ/mol) Ag(s) 0 HNO3(aq) -207 AgNO3(aq) -101 NO(g) 90 H2O(l) -286 Based on your answers to parts (a) and (b), is the reaction more likely to be thermodynamically favorable at 25°C or at 95°C? Justify your answer. The student runs the reaction using a 3 to 4 mole ratio of Ag(s) to HNO3(aq). Suggest a method the student can use to isolate solid AgNO3 from the other products of the reaction.

The Correct Answer and Explanation is:

a. Entropy Change (ΔS):

To predict the sign of the entropy change (ΔS) for the reaction, we need to consider the phases and the number of molecules involved in the reactants and products.

The reaction is:
Ag(s) + 4 HNO3(aq) → 3 AgNO3(aq) + NO(g) + 2 H2O(l)

• On the left side (reactants), we have:
  • Ag(s) (solid): low entropy (solids generally have lower entropy than liquids or gases).
  • HNO3(aq) (aqueous): moderate entropy.
• On the right side (products), we have:
  • AgNO3(aq) (aqueous): moderate entropy.
  • NO(g) (gas): high entropy (gases have the highest entropy).
  • H2O(l) (liquid): lower entropy than gases but higher than solids.

Entropy tends to increase when there is a phase change from solid to liquid or gas, or when the number of gas molecules increases. In this reaction:

• The number of moles of gas increases from 0 (on the left) to 1 (on the right).
• The formation of NO(g), a gas, suggests an increase in disorder.

Thus, ΔS is positive, meaning there is an increase in entropy due to the production of a gas and the phase changes.

b. Standard Enthalpy Change (ΔHrxn):

We can calculate the standard enthalpy change (ΔHrxn) using the formula:ΔHrxn=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_{\text{rxn}} = \sum \Delta H_f^\circ (\text{products}) – \sum \Delta H_f^\circ (\text{reactants})ΔHrxn​=∑ΔHf∘​(products)−∑ΔHf∘​(reactants)

From the table:

• ΔHf° for Ag(s) = 0 kJ/mol
• ΔHf° for HNO3(aq) = -207 kJ/mol
• ΔHf° for AgNO3(aq) = -101 kJ/mol
• ΔHf° for NO(g) = 90 kJ/mol
• ΔHf° for H2O(l) = -286 kJ/mol

Now, calculate:ΔHrxn=[3×(−101)+1×90+2×(−286)]−[1×0+4×(−207)]\Delta H_{\text{rxn}} = [3 \times (-101) + 1 \times 90 + 2 \times (-286)] – [1 \times 0 + 4 \times (-207)]ΔHrxn​=[3×(−101)+1×90+2×(−286)]−[1×0+4×(−207)]ΔHrxn=[−303+90−572]−[0−828]\Delta H_{\text{rxn}} = [-303 + 90 – 572] – [0 – 828]ΔHrxn​=[−303+90−572]−[0−828]ΔHrxn=−785−(−828)=−785+828=43 kJ/mol\Delta H_{\text{rxn}} = -785 – (-828) = -785 + 828 = 43 \, \text{kJ/mol}ΔHrxn​=−785−(−828)=−785+828=43kJ/mol

Therefore, the standard enthalpy change (ΔHrxn) for the reaction is +43 kJ/mol.

c. Thermodynamic Favorability at 25°C vs 95°C:

To determine whether the reaction is more likely to be thermodynamically favorable at 25°C or 95°C, we need to consider the relationship between enthalpy (ΔH) and entropy (ΔS) changes. The Gibbs free energy (ΔG) determines whether a reaction is thermodynamically favorable:ΔG=ΔH−TΔS\Delta G = \Delta H – T\Delta SΔG=ΔH−TΔS

At 25°C (298 K):

• ΔH = +43 kJ/mol (positive, endothermic)
• ΔS is positive, so at a lower temperature, the TΔS term is smaller.
• Therefore, ΔG may be positive, indicating that the reaction is less likely to be spontaneous at lower temperatures.

At 95°C (368 K):

• The temperature is higher, and TΔS increases, which makes the ΔG more negative (because TΔS becomes larger than ΔH).
• Hence, the reaction is more likely to be thermodynamically favorable at 95°C.

d. Method to Isolate Solid AgNO3:

To isolate solid AgNO3 from the other products, the student can:

1. Evaporate the water: Heat the reaction mixture gently to remove the water (H2O) and obtain AgNO3 as a solid since it is highly soluble in water and will remain in the aqueous solution.
2. Filter the NO(g): Since NO is a gas, it will escape from the solution, leaving behind the aqueous AgNO3.
3. Crystallize AgNO3: Once the solution is concentrated, the student can cool it to allow the AgNO3 to crystallize out of the solution, then filter it off to isolate the solid AgNO3.