Fe3O4(s) + 4H2(g)3Fe(s) + 4H2O(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.79 moles of Fe3O4(s) react at standard conditions. S°surroundings = J/K

dS= 1.79*169.504

j/k = 303.41 j/k

Explanation:

Fe3O4(s) + 4H2(g) --> 3Fe (s)+ 4H2O(g)

dS(Fe3O4) =146.4 j/k

dS(H2) =130.684

dS(Fe) =27.78

dS(H2O) =188.825

dSrxn = dS[product]-dS[reactants]

= 3*dS(Fe)+ 4*dS(H2O)-[1*dS(Fe3O4)+ 4dS(H2)]

= [3*27.78 +4*188.825-146.4 -4*130.684] j/k = 169.504 j/k

This is the dS for 1mole Fe3O4

for 1.79 mols Fe3O4

dS= 1.79*169.504 j/k = 303.41 j/k

the earth would become to overheated and everything alive on it will eventually die from overheating. plants will eventually die as well, water will evaporate, and it will be nothing but a hot planet without anything living on it.