Copper chloride ...............
Na₂CrO₄ + PbCl₂ → PbCrO₄ + 2 NaCl
<u>Explanation:</u>
In a double displacement reaction, the reactants which are involved in the reaction exchanging their ions thereby produces 2 new compounds. Here sodium chromate and lead chloride are undergoing double displacement reaction, the ions exchanges their position there by forming sodium chloride and lead chromate. So the double displacement reaction is given as,
Na₂CrO₄ + PbCl₂ → PbCrO₄ + 2 NaCl
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
