For the reaction 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g) H° = -6.0 kJ and S° = 88.7...

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For the reaction 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g) H° = -6.0 kJ and S° = 88.7...

For the reaction

3Fe₂O₃(s) + H₂(g)2Fe₃O₄(s) + H₂O(g)

H° = -6.0 kJ and S° = 88.7 J/K

The equilibrium constant for this reaction at 278.0 K is

Assume that H° and S° are independent of temperature.

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Answer 1

Answer #1

3F (8) + 14 15) 2F0₂84151 +1.2015). H -6.0kJ 450 = 88.79 cien tenparalone T-278.0K A6 Alt - TASO =-6.0kg - (278.0kx 88. 7x10k

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Answer 2

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For the reaction 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g) H° = -6.0 kJ and S° = 88.7...

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Homework Answers

Add Answer to: For the reaction 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g) H° = -6.0 kJ and S° = 88.7...

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For the reaction 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g) H° = -6.0 kJ and S° = 88.7...