ThermodynamicsHard
Question
Benzene burns according to the following equation at 300 K (R = 8.314 J mole-1K-1)
2C6H6(l) + 15 O2 (g) → 12 CO2 (g) + 6H2O(l) ᐃHo = - 6542 kJ/mol
What is the ᐃEo for the combustion of 1.5 mol of benzene
2C6H6(l) + 15 O2 (g) → 12 CO2 (g) + 6H2O(l) ᐃHo = - 6542 kJ/mol
What is the ᐃEo for the combustion of 1.5 mol of benzene
Options
A.-3271 kJ
B.-9813 kJ
C.-4906.5 kJ
D.None of these
Solution
From given reaction ᐃng = 12 - 15 = - 3
so ᐃEo = ᐃHo - ᐃng RT = - 6542 + 3RT
for 1.5 mole, ᐃEo =
{-6542 + 3RT} = 4912.11 KJ
so ᐃEo = ᐃHo - ᐃng RT = - 6542 + 3RT
for 1.5 mole, ᐃEo =
{-6542 + 3RT} = 4912.11 KJ Create a free account to view solution
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