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Answer:
5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄+ 2H₂O
Explanation:
PbO₂ + MnSO₄ + HNO₃ ⟶ HMnO₄ + PbSO₄ + Pb(NO₃)₂ + H₂O
It will be easiest to balance this equation by the ion-electron method.
1. Write the ionic equation
PbO₂ + Mn²⁺ + SO₄²⁻ + H⁺ + NO₃⁻ ⟶ H⁺ + MnO₄⁻ + Pb²⁺ + SO₄²⁻ + Pb²⁺ + NO₃⁻ + H₂O
2. Eliminate H⁺, H₂O, and spectator ions
PbO₂ + Mn²⁺ ⟶ MnO₄⁻ + Pb²⁺
3. Separate the skeleton equation into two half-reactions.
PbO₂ ⟶ Pb²⁺
Mn²⁺ ⟶ MnO₄⁻
4. Balance all atoms other than H and O
Done
5. Balance O by adding water molecules to the deficient side
PbO₂ ⟶ Pb²⁺ + 2H₂O
Mn²⁺ + 4H₂O ⟶ MnO₄⁻
6. Balance H by adding H⁺ ions to the deficient side.
PbO₂+ 4H⁺ ⟶ Pb²⁺ + 2H₂O
Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺
7. Balance charge by adding electrons to the deficient side.
PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O
Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e-
8. Multiply each half-reaction by a number to equalize the electrons transferred.
5 × [PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O]
2 × [Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e⁻]
9. Add the two half-reactions.
5PbO₂+ 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 10H₂O
<u> 2Mn²⁺ + 8H₂O ⟶ 2MnO₄⁻ + 16H⁺ + 10e⁻ </u>
5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻
10. Cancel species that occur on each side of the equation
5PbO₂ + 2Mn² + <u>8H₂O</u> + <u>20H⁺</u> + <u>10e⁻</u> ⟶ 5Pb²⁺ + 2MnO₄⁻ + <u>10H₂O</u> + <u>16H⁺</u> + <u>10e⁻ </u>
becomes
5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O
11. Add the missing spectator ions
5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O
+ 2SO₄²⁻ + 4NO₃⁻ + 2H⁺ + 2NO₃⁻ +2SO₄²⁻ + 6NO₃⁻ + 2H⁺
becomes
5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O
12. Check that all atoms are balanced.
Everything checks. The balanced equation is
5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O
Explanation:
- For path A, the calculation will be as follows.
As, for reversible isothermal expansion the formula is as follows.
W =
Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.
As the given data is as follows.
R = 8.314 J/(K mol), T = 298 K ,
= 8.34 L,
= 2.67 L
Now, putting the given values into the above formula as follows.
W =
=
=
= -2818.68 J
Hence, work for path A is -2818.68 J.
- For path B, the calculation will be as follows.
Step 1: When there is no change in volume then W = 0
Hence, for step 1, W = 0
Step 2: As, the gas is allowed to expand against constant external pressure = 1.00 atm.
So, W =
Now, putting the given values into the above formula as follows.
W =
=
= -5.67 atm L
As we known that, 1 atm L = 101.33 J
Hence, work will be calculated as follows.
W =
= -574.54 J
Therefore, total work done by path B = 0 + (-574.54 J)
W = -574.54 J
Hence, work for path B is -574.54 J.