In: Chemistry
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
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