Answer to Question #349390 in General Chemistry for rcy

Question

Calculate the standard entropy changes for the following reactions at 25°C:

a) H2(g) + CuO(s) → Cu(s) + H2O(g)

b) 2Al(s) + 3ZnO(s) → Al2O3(s) + 3Zn(s)

c) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

d) S(s) + O2(g) → SO2(g)

e) MgCO3(s) → MgO(s) + CO2(g)

Answer 1

∆S = S_produced - S_reactant

a.)

H₂ + CuO = Cu + H₂O

S (H₂) = 130,7 J/(mol*K)

S (CuO) = 42,6 J/(mol*K)

S (Cu) = 33,17 J/(mol*K)

S (H₂O) = 188,7 J/(mol*K)

∆S = (33,17+188,7) - (130,7+42,6) = 48,57 J/(mol*K)

b.)

2Al + 3ZnO = Al₂O₃ + 3Zn

S(Al) = 28,27 J/(mol*K)

S(ZnO) = 43,65 J/(mol*K)

S(Al₂O₃) = 50,92 J/(mol*K)

S(Zn) = 41,72 J/(mol*K)

∆S = (50,92+ 3*41,72) - (2*28,27+3*43,65) = -11,41 J/(mol*K)

c.)

CH₄ + 2O₂ = CO₂ + 2H₂O

S(CH₄) = 186,3 J/(mol*K)

S(O₂) = 205 J/(mol*K)

S(CO₂) = 213,79 J/(mol*K)

S(H₂O) = 188,7 J/(mol*K)

∆S = (213,79+2*188,7) - (186,3+2*205) = -5,11 J/(mol*K)

d.)

S + O₂ = SO₂

S(S) = 32,05 J/(mol*K)

S(O₂) = 205 J/(mol*K)

S(SO₂) = 248,22 J/(mol*K)

∆S = 248,22 - (32,05+205) = 11,17 J/(mol*K)

e.)

MgCO₃ = MgO + CO₂

S(MgCO₃) = 65,84 J/(mol*K)

S(MgO) = 26,9 J/(mol*K)

S(CO₂) = 213,79 J/(mol*K)

∆S = 26,9 + 213,79 - 65,84 = 174,84 J/(mol*K)