at 300 k the standard enthalpies of formation of c6h5cooh ,co2 and h2o are -408 ,-393 and -286 kj mol-1 respectively calculate heat of combustion of benzoic acid at (1) constant pressure (2) constant volume
Answers
E = -321.30 KJ = - 321.30 X1000 = -321300J
C6H5COOH + 15/2 O2 → 7CO2 + 3 H2O
∆ n = 10-(1+15/2 )
= 10-17/2
= 1.5 moles
T = 273 + 27 = 300K
R = 8.314 JK-1mol-1
The heat of combustion at constant pressure is
∆H = ∆E + ∆nRT
=-321300 + 1.5 X 8.314 X 300
= - 317565J
The heat of combustion at constant pressure = - 317565J =-317.565 KJ
Answer:
Benzoic acid burns at a temperature of 75KJ/mol.
Explanation:
What is the heat of combustion of benzoic acid ?
E = -321.30 KJ = - 321.30 X1000 = -321300J
C6H5COOH + 15/2 O2 → 7CO2 + 3 H2O
∆ n = 10-(1+15/2 )
= 10-17/2
= 1.5 moles
T = 273 + 27 = 300K
R = 8.314 JK-1mol-1
The heat of combustion at constant pressure is
∆H = ∆E + ∆nRT
=-321300 + 1.5 X 8.314 X 300
= - 317565J
Heat of combustion at constant pressure = - 317.565 J = - 317.565 KJ
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