<span>water vapour.
carbon dioxide.
carbon monoxide.particles.
<span>sulfur dioxide.</span></span>
Answer:
ΔS° = 180.5 J/mol.K
Explanation:
Let's consider the following reaction.
4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)
The standard molar entropy of the reaction (ΔS°) can be calculated using the following expression.
ΔS° = ∑np × S°p - ∑nr × S°r
where,
ni are the moles of reactants and products
S°i are the standard molar entropies of reactants and products
ΔS° = 4 mol × S°(NO(g)) + 6 × S°(H₂O(g)) - 4 mol × S°(NH₃(g)) - 5 mol × S°(O₂(g))
ΔS° = 4 mol × 210.8 J/K.mol + 6 × 188.8 j/K.mol - 4 mol × 192.5 J/K.mol - 5 mol × 205.1 J/K.mol
ΔS° = 180.5 J/K
This is the change in the entropy per mole of reaction.
Base on your question where a concentration cell consist of two SN/SN2+half cells. The solution in one half cell A is 0.13M SN(NO3)2 and is 0.87 M Sn(NO3)2 in the other half cell to get the cell potential at 25 degree the answer is 0.059/2 log0.13/0.87
