Questions: Given the table of standard molar entropy values of heat substance in the reaction below, what is the standard change in entropy for the reaction? 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g)

Solution Steps

• Identify the reactants and products in the given chemical reaction:
  • Reactants: \(4 \text{NH}_3(g)\) and \(3 \text{O}_2(g)\)
  • Products: \(2 \text{N}_2(g)\) and \(6 \text{H}_2\text{O}(g)\)

• The standard change in entropy (\(\Delta S^\circ\)) for a reaction is calculated using the formula: \[ \Delta S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}} \]

• Calculate the total standard molar entropy for the products: \[ \text{Total } S^\circ_{\text{products}} = (2 \times S^\circ_{\text{N}_2(g)}) + (6 \times S^\circ_{\text{H}_2\text{O}(g)}) \] \[ = (2 \times 211) + (6 \times 189) \]

• Calculate the total standard molar entropy for the reactants: \[ \text{Total } S^\circ_{\text{reactants}} = (4 \times S^\circ_{\text{NH}_3(g)}) + (3 \times S^\circ_{\text{O}_2(g)}) \] \[ = (4 \times 193) + (3 \times 205) \]

• Substitute the calculated values into the entropy change formula: \[ \Delta S^\circ = \left[(2 \times 211) + (6 \times 189)\right] - \left[(4 \times 193) + (3 \times 205)\right] \]

Final Answer