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rusak2 [61]
3 years ago
9

What is the net ionic equation for the precipitation reaction between BaCl2 and Na2So4? I know the answer ( Ba2+(aq) + SO42-(aq)

--> BaSO4 (S) ), but i dont understand how it is the answer. Can anybody help? <3
Chemistry
1 answer:
elixir [45]3 years ago
5 0

Answer:

Net ionic equation:

Ba²⁺(aq) + SO₄²⁻(aq)  →   BaSO₄(s)

Explanation:

Chemical equation:

BaCl₂ +  Na₂SO₄   →   BaSO₄ + NaCl

Balanced Chemical equation:

BaCl₂(aq) +  Na₂SO₄(aq)   →   BaSO₄(s) + 2NaCl(aq)

Ionic equation:

Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq)  →  BaSO₄(s)+ 2Na⁺(aq) + 2Cl⁻ (aq)

Net ionic equation:

Ba²⁺(aq) + SO₄²⁻(aq)  →   BaSO₄(s)

The Cl⁻(aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The BaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.  

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Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ
JulsSmile [24]

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

\frac{1}{2}Br2(g) ⇒ Br(g) ,  Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) ,  Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

   average bond enthalpy is \frac{1134.9}{4} = 283.725 kJ ⋅ mol − 1

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