Calculate ΔrG° for the reaction below at
25.0 °C
CH4(g) + H2O(g) → 3 H2(g) +
CO(g)
given ΔfG° [CH4(g)] = –50.8 kJ/mol,
ΔfG° [H2O(g)] = –228.6 kJ/mol,
ΔfG° [H2(g)] = 0.0 kJ/mol, and
ΔfG° [CO(g)] = –137.2 kJ/mol.
ΔrG° reaction = ΔrG° products - ΔrG° reactants
= ( 3 x 0 -137.2) - ( -50.8 -228.6)
= -137.2 + 279.4
= 142.2 kJ / mol
ΔrG° reaction = 142.2 kJ / mol
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