Question

Given the following, determine ΔG°f at 298 K for SnO. Sn(s) + SnO2(s) → 2SnO(s) ; ΔG° = 12.0 kJ at 298K

Substance ΔG°f (kJ/mol) at 298 K

SnO( s) ?

SnO 2( s) –515.8

Answer 1

Answer:

The value of change in Gibbs free energy for tin(II) oxide solid is -251.9 kJ/mol.

Explanation:

$Sn(s) + SnO_2(s)\rightarrow 2SnO(s), \Delta G_{f}^{o} = 12.0 kJ$

$\Delta G_{f,SnO_2}^{o}= -515.8kJ/mol$

$\Delta G_{f,Sn}^{o}= 0 kJ/mol$

$\Delta G_{f,SnO}^{o}=?$

$\Delta G_{f}^{o}=\sum ((\Delta G_{f}^{o})_ {products})-\sum ((\Delta G_{f}^{o})_ {reactants})$

$12.0 kJ=(2 mol\times \Delta G_{f,SnO}^{o})-(1mol\times 0 kJ/mol+1 mol\times -515.8 kJ/mol)$

$\Delta G_{f,SnO}^{o}=-251.9 kJ/mol$

The value of change in Gibbs free energy for tin(II) oxide solid is -251.9 kJ/mol.