ThermodynamicsHard
Question
The difference between heats of reaction at constant pressure and constant volume for the reaction
2C6H6(l) + 1502(g) → 12CO2(g) + 6H2 O(l) at 25oC in kJ is :
2C6H6(l) + 1502(g) → 12CO2(g) + 6H2 O(l) at 25oC in kJ is :
Options
A.- 7.43
B.+3.72
C.- 3.72
D.+7.43
Solution
Heat of reaction at constant pressure = ᐃH and at
constant volume = ᐃH
But ᐃH = ᐃU + ᐃnRT
Fro the reaction
2C6H6(l) + 15O2(g) → 12CO2(g) + 6h2O(l)
ᐃn = 12 - 15 = - 3 (Count gasecos molecules only)
and ᐃH - ᐃU = ᐃnRT
= - 3 × 8.314 × (273 + 25)
= - 74327 J ⇒ - 7.43KJ
constant volume = ᐃH
But ᐃH = ᐃU + ᐃnRT
Fro the reaction
2C6H6(l) + 15O2(g) → 12CO2(g) + 6h2O(l)
ᐃn = 12 - 15 = - 3 (Count gasecos molecules only)
and ᐃH - ᐃU = ᐃnRT
= - 3 × 8.314 × (273 + 25)
= - 74327 J ⇒ - 7.43KJ
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