2Al + Cr2O3 → 2Cr + Al2O3
ΔHr0 = [2ΔHf (Cr) + ΔHf (Al2O3)]–
[2ΔHf (Al) + ΔHf
(Cr2O3)] ΔHr0 = [0 + (–
1596 kJ)]– [0 + (– 1134)]
ΔHr0 = – 1596
kJ + 1134 kJ
ΔHr0 = – 462
kJ
ΔU = q + w
ΔU = – 1 kJ + 4 kJ
ΔU = + 3kJ
Fe + 2HCl → FeCl2 + H2
1 mole of
Iron liberates 1 mole of
Hydrogen gas
55.85 g Iron
= 1 mole Iron
∴ n = 1
T = 25° C =
298 K
w = – P Δ V
w =-P (nRT/P)
w = – nRT
w = –1 ×
8.314 × 298 J
w = –
2477.57 J
w = – 2.48
kJ
Ti = 125° C = 398 K
Tf = 25° C = 298 K
ΔH = nCp (Tf – Ti)
ΔH = 2x(5/2)R(298-398)
ΔH = – 500 R
Given :
ΔHC (CH4)= – 890 kJ
mol–1
ΔHC (C3H8)=
– 2220 kJ mol–1
4EC–H= 360 kJ mol–1
EC–H= 90 kJ mol–1
EC–C + 6 EC–H = 620 kJ mol–1
EC–C + 6 × 90 = 620 kJ mol–1
EC–C + 540= 620 kJ mol–1
EC–C= 80 kJ mol–1