What is the free energy change for the reaction SiO₂(s) + Pb(s) $\rarr$ PbO₂(s) + Si(s)
$\Delta$ G^$\circ$_f (PbO₂) = -217 kJ/mol
$\Delta$ G^$\circ$_f (SiO₂) = -856 kJ/mol

Question

What is the free energy change for the reaction SiO₂(s) + Pb(s)

\rarr

PbO₂(s) + Si(s)

\Delta

G^$\circ$_f (PbO₂) = -217 kJ/mol

\Delta

G^$\circ$_f (SiO₂) = -856 kJ/mol

Quizplus · Accepted Answer

The Answer of What is the free energy change for the reaction SiO₂(s) + Pb(s) $\rarr$PbO₂(s) + Si(s) $\Delta$G^$\circ$_f (PbO₂) = -217 kJ/mol $\Delta$G^$\circ$_f (SiO₂) = -856 kJ/mol A) 619 kJ/mol B) 639 kJ/mol C) 659 kJ/mol D) 679 kJ/mol E) None of the above

What Is the Free Energy Change for the Reaction SiO2(s) →\rarr→

What Is the Free Energy Change for the Reaction SiO₂(s) $\rarr$