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 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 Delta_(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 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 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?

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)

The heat of formation of Fe_(2)O_(3) is -824.2 kJ mol^(-1). Delta H for the reaction 2Fe_(2)O_(3)(s)rarr 4Fe(s)+3O_(2)(g) is :

the standard heat of formation of Fe2O3 (s) is 824.2kJ mol-1 Calculate heat change for the reaction. 4Fe(s) + 302 (g)= 2Fe2O3(s)