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207
2SO 2 (g) + O 2 (g) ⎯⎯
→ 2SO 3 (g)
Decomposition of 2 moles of NH4NO3 at
100ºC.
∴
∴
= –8.314 (JK–1 mol–1)
× 373 (K) × (6 – 0) (mol)
= –18610 J
= –18.610 kJ
W = –18.61 kJ
Work is done by the system
NH 4 NO 3 (s) ⎯⎯
→ N 2 O(g) + 2H 2 O(g)
Given :
1
(a) one mole of SO2 reacts with mole of
2
O2 to form 1mole of SO3
Hence, n1 = 1.5 mol, n2 = 1mole,
R = 8.314 JK–1 mol–1
T = 50ºC + 273 = 323 K
To find :
W=?
Solution :
W = –Δn RT
= –RT (n2 – n1)
= –8.314 (JK–1mol–1) × 323 (K) × (1 – 1.5) (mol)
= –8.314 × 323 × (–0.5) J
∴ W = +1343 J
∴ Work is done on the system.
*7.
The enthalpy change for the reaction
*6.
(a)
(b)
Calculate the work done in each of the
following reactions. State whether work is
done on or by the system.
The oxidation of one mole of SO2 at 50ºC.
Given :
(b) 2 moles of NH4NO3(s) gives 2 moles of
N2O(g) and 4 moles of H2O(g)
Hence, n1 = 0,
n = 6,
R = 8.314 JK–1 mol–1
T = 100ºC + 273K = 373K.
To find :
W=?
Solution :
W = –RT (n2 – n1)
C 2 H 4 (g)+H 2 (g) ⎯⎯
→ C 2 H 6 (g) is –620J
when 100 mL of ethylene and 100 mL of
H2 react at 1 atm pressure. Calculate the
pressure-volume work and ΔU.
Given :
Initial volume (V1) = 100 mL + 100 mL
= 200 mL.
Final volume (V2) = 100 mL, Δ H = –620J
∴ ΔV = V2 – V1
= 100 – 200
= –100 mL
= –0.1 L
To find :
Work = ? and ΔU = ?
Solution :
W = –p ex ⋅ ΔV
= –1 atm × (–0.1L)
= +0.1 L atm.
But, 1 L atm = 101.3 J
∴
W = +0.1 (L.atm) × 101.3
J
L ⋅ atm
W = +10.13 J
We know, ΔH = ΔU + P ⋅ ΔV
or ΔU = ΔH – P ⋅ ΔV
= –620J – (–10.13 J)
ΔU = –609.87 J.
Chapter - 3 Chemical Thermodynamics And Energetics
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