Consider the following reaction at 298 K: 4Al(s) + 3O2(g) ==> 2Al2O3(s) Delta H= -3351.4 kJ/mol...

Consider the following reaction at 298 K: 4Al(s) + 3O2(g) ==> 2Al2O3(s)
Delta H= -3351.4 kJ/mol

Calculate:

a. Delta Ssystem = _______J/mol*K

b. Delta Ssurroundings = _______J/mol*K

c. Delta S universe = _________J/mol*K

Solutions

Expert Solution

delta Sf (AL2O3) = 50.92 J/molK
delta Sf (O2) = 205.14 J/molK
delta Sf (Al) = 28.33 J/molK

Reaction taking place is:
4Al +3 O2 --->2Al2O3
a.
delts S sys = delta Sf (products) - delta Sf (reactsnta)
                      = 2*delta Sf (Al2O3) - 4* delta Sf (Al) - 3*delta Sf (O2)
                      = 2*(50.92) - 4*(28.33) -3*(205.14)
                       = -626.9 J/molK
b.
q sys= delta H = -3351.4 KJ/mol = -3351400 J/mol
q surr = -qsys = 3351400 J/mol
I am assuming tempearture os 298 K
delta S surrounding = qsurr/T
                                           =3351400/298
                                            = 11246.3 J/molK

c.
delta S univ = delta S sys +delta S surr
= -626.9 +11246.31
= 10619.4 J/molK

queen_honey_blossom answered 5 months ago

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