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3Fe(s) + 4H2O(g) ⇄ Fe3O4(s) + 4H2(g). The equilibrium constant, K, of the reaction above is represented as
\(\dfrac{[\mathrm{Fe_3O_4}][\mathrm{H_2}]}{([\mathrm{Fe}][\mathrm{H_2O}])}\)
\(\dfrac{[\mathrm{H_2O}]^{4}}{[\mathrm{H_2}]^{4}}\)
\(\dfrac{[\mathrm{H_2}]^{4}}{[\mathrm{H_2O}]^{4}}\)
\(\dfrac{[\mathrm{Fe}]^{3}[\mathrm{H_2O}]^{2}}{([\mathrm{Fe_3O_4}][\mathrm{H_2}]^{4})}\)
100 kJ
300 kJ
250 kJ
200 kJ
The minimum amount of energy required for a reaction to take place is
Lattice energy
Ionization energy
Activation energy
Kinetic energy
2SO2(g) + O2(g) ⇄ 2SO3(g), ΔH = −395.7 kJ mol−1. In the reaction above, the concentration of SO3(g) can be increased by
Decreasing the pressure
Decreasing the temperature
Increasing the temperature
Adding a catalyst
In an endothermic reaction, if there is a loss in entropy the reaction will
Be indeterminate
Be spontaneous
Not be spontaneous
Be at equilibrium
