Use appendix C to calculate ∆So for the following a) CH4 (g) + 2
O2 (g) → CO2 (g) + 2 H2O (g)
b) 6 CO2 (g) + 6 H2O (l) → C6H12O6 (s) + 6 O2 (g)
c) P4O10 (s) + 6 H2O (l) → 4 H3PO4 (aq)
d) NH3 (g) + HCl (g) → NH4Cl (s)
dS:
a) CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g)
dS = Sproducts - Sreactants
dS = (CO2 + 2H2O) -(CH4+2O2)
dS = (213.8+ 2*188.8) -(186.3+2*205.2)
dS= -5.3 J/molK
b) 6 CO2 (g) + 6 H2O (l) → C6H12O6 (s) + 6 O2 (g)
dS = Sproducts - Sreactants
dS = (C6H12O6 + 6*O2) - (6*CO2 + 6*H2O)
dS = (C6H12O6 + 6*O2) - (6*213.8+ 6*70.0)
dS = (212+ 6*205.5) - (6*213.8+ 6*70.0)
dS = -257.8 J/molK
c) P4O10 (s) + 6 H2O (l) → 4 H3PO4 (aq)
dS = Sproducts - Sreactants
dS = (4*H3PO4) - (P4O10 + 6*H2O )
dS = (4*158.2 ) - (228.86+ 6*70.0)
dS = -16.06 J/molK
d) NH3 (g) + HCl (g) → NH4Cl (s)
dS = Sproducts - Sreactants
dS = (NH4Cl (s) ) - (NH3 (g) + HCl (g) )
dS = (94.6) - (192.8 + 186.9)
dS = -285.1 J/Kmol
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