A student determines the value of the equilibrium constant to be 1.03×106 for the following reaction. 3Fe2O3(s) + H2(g)
Posted: Sun May 22, 2022 8:42 pm
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
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