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The heat of formation of Fe(2)O(3)(s) is...

The heat of formation of `Fe_(2)O_(3)(s)` is -824.0 kJ.What will be the `Delta H` for thereaction:
`2Fe_(2)O_(3)(s) to 4Fe(s) + 3O_(2)(g)`

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To find the change in enthalpy (ΔH) for the reaction: \[ 2 \text{Fe}_2\text{O}_3(s) \rightarrow 4 \text{Fe}(s) + 3 \text{O}_2(g) \] we start with the given heat of formation for iron(III) oxide (Fe₂O₃): 1. **Understanding Heat of Formation**: The heat of formation (ΔH_f) of Fe₂O₃(s) is given as -824.0 kJ. This value indicates that when 1 mole of Fe₂O₃ is formed from its elements (iron and oxygen), 824.0 kJ of energy is released. ...
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The enthalpy of formation of Fe_(2)O_(3)(s) is -824.2 kJ "mol"^(-1) .Calculate the enthalpy change for the reaction : 4Fe(s) + 3O_(2)(g) to 2Fe_(2)O_(3)(s)

Delta_(f)H^(@) of Fe_(2)O_(3) is -825 kJ. What is the enthalpy change for the reaction? 4Fe_(2)O_(3) to 8Fe + 6O_(2)

What is the value of Delta s^(@) for the reaction below? Fe_(2)O_(3)(s)+3CO(g)to2Fe(s)+3CO_(2)(g)

The resting of iron occurs as: 4Fe(s) +3O_(2)(g) rarr 2Fe_(2)O_(3)(s) The entalpy of formation of Fe_(2)O_(3)(s) is -824.0 kJ mol^(-1) and entropy change for the reaction is +550 J K^(-1) mol^(-1) . Calculate Deta_(surr)S and predict whether resuting of iron is spontaneous or not at 298 K . Given Delta_(sys) H =- 553.0 J K^(-1) mol^(-1)

The rusting of iron occurs as 4Fe(s)+3O_(2)(g)to 2Fe_(2)O_(3)(s) . Enthalpy of formation of Fe_(2)O_(3)(s) is -824.2 kJ mol^(-1) and entropy change for the reaction, i.e., Delta S_("system") , is -549 J K^(-1) mol^(-1) . Calculate Delta S_("surrounding") and predict whether rusting of iron is spontaneous or not at 298 K.

The following two reactions are known: Fe_(2)O_(3)(s)+3CO(g)to 2Fe(s)+3CO_(2)(g),DeltaH=-26.8kJ FeO(s)+CO(g) to Fe(s)+CO_(2)(g),DeltaH=-16.5kJ The value of DeltaH for the following reaction Fe_(2)O_(3)+CO(g) to 2FeO(s)+CO_(2)(g) is

The oxidation of iron occurs as: 4Fe(s) + 3 O_(2)(g) to 2Fe_(2)O_(3)(s) The enthalpy of formation of Fe_(2)O_(3) is - 824.2 kJ mol^(-1) and entropy change for the reaction is -549 J K^(-1) mol^(-1) at 298 K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous at 298 K?

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