A student determines the value of the equilibrium constant to be 1.03×106 for the following reaction. 3Fe2O3(s) + H2(g)

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answerhappygod
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A student determines the value of the equilibrium constant to be 1.03×106 for the following reaction. 3Fe2O3(s) + H2(g)

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A student determines the value of the equilibrium constant to
be 1.03×106 for the following
reaction.

3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g)

Based on this value of Keq:

G° for this
reaction is expected to be (greater, less) fill in the blank
1 than zero.

Calculate the free energy change for the reaction
of 1.67 moles
of Fe2O3(s) at
standard conditions at 298K.
G°rxn = kJ
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