Question

"Thermite" reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is 3 Mg(s) + Cr2O3(s) → 3 MgO(s) + 2 Cr(s). During the reaction, the surroundings absorb 665.1 kJ/mol of heat. Is the reaction spontaneous at room temperature (298.15 K) under standard conditions?

Answer 1

Answer: The given reaction is non-spontaneous in nature.

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]$

For the given chemical reaction:

$3Mg(s)+Cr_2O_3(s)\rightarrow 3MgO(s)+2Cr(s)$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(3\times \Delta S^o_{(MgO(s))})+(2\times \Delta S^o_{(Cr(s))})]-[(3\times \Delta S^o_{(Mg(s))})+(1\times \Delta S^o_{(Cr_2O_3(s))})]$

We are given:

$\Delta S^o_{(Mg(s))}=32.68J/K.mol\\\Delta S^o_{(Cr_2O_3(s))}=81.2J/K.mol\\\Delta S^o_{(MgO(s))}=26.94J/K.mol\\\Delta S^o_{(Cr(s))}=23.77J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(3\times (26.94))+(2\times (23.77))]-[(3\times (32.68))+(1\times (81.2))]\\\\\Delta S^o_{rxn}=-50.88J/K=-0.0509kJ/K.mol$

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

To calculate the standard Gibbs free energy of the reaction, we use the equation:

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy change of the reaction = 665.1 kJ/mol

T = Temperature = 298.15 K

$\Delta S^o$ = standard entropy change of the reaction = -0.0509 kJ/K.mol

Putting values in above equation, we get:

$\Delta G^o=(665.1kJ/mol)-(298.15K\times (-0.0509kJ/K.mol))=680.27kJ/mol$

As, the Gibbs free energy of the reaction is coming out to be positive, the reaction is non-spontaneous in nature.

Hence, the given reaction is non-spontaneous in nature.