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Transcript
```207
2SO 2 (g) + O 2 (g) â¯â¯
â 2SO 3 (g)
Decomposition of 2 moles of NH4NO3 at
100ÂºC.
â´
â´
= â8.314 (JKâ1 molâ1)
Ã 373 (K) Ã (6 â 0) (mol)
= â18610 J
= â18.610 kJ
W = â18.61 kJ
Work is done by the system
NH 4 NO 3 (s) â¯â¯
â 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 JKâ1 molâ1
T = 50ÂºC + 273 = 323 K
To find :
W=?
Solution :
W = âÎn RT
= âRT (n2 â n1)
= â8.314 (JKâ1molâ1) Ã 323 (K) Ã (1 â 1.5) (mol)
= â8.314 Ã 323 Ã (â0.5) J
â´ W = +1343 J
â´ 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 50ÂºC.
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 JKâ1 molâ1
T = 100ÂºC + 273K = 373K.
To find :
W=?
Solution :
W = âRT (n2 â n1)
C 2 H 4 (g)+H 2 (g) â¯â¯
â C 2 H 6 (g) is â620J
when 100 mL of ethylene and 100 mL of
H2 react at 1 atm pressure. Calculate the
pressure-volume work and ÎU.
Given :
Initial volume (V1) = 100 mL + 100 mL
= 200 mL.
Final volume (V2) = 100 mL, Î H = â620J
â´ ÎV = V2 â V1
= 100 â 200
= â100 mL
= â0.1 L
To find :
Work = ? and ÎU = ?
Solution :
W = âp ex â ÎV
= â1 atm Ã (â0.1L)
= +0.1 L atm.
But, 1 L atm = 101.3 J
â´
W = +0.1 (L.atm) Ã 101.3
J
L â atm
W = +10.13 J
We know, ÎH = ÎU + P â ÎV
or ÎU = ÎH â P â ÎV
= â620J â (â10.13 J)
ÎU = â609.87 J.
Chapter - 3 Chemical Thermodynamics And Energetics
```
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