What is ∆G° at 298 K for the disproportionation of PbO(s)? 2 PbO(s) ® PbO2(s) + Pb(s) ∆G° = ??? | Electrochemistry

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Question

What is ∆G° at 298 K for the disproportionation of PbO(s)?
2 PbO(s) ® PbO₂(s) + Pb(s) ∆G° = ???

163 kJ mol^-1

✓ Correct

241 kJ mol^-1

403 kJ mol^-1

483 kJ mol^-1

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