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What is the difference between ΔH and ΔE (on a molar basis) for the combustion of n-octane (l) at 25 degree C°?
one month ago

Answers : (2)

Arun
22809 Points
							
Dear student
 
∆H – ∆E = – 4.5 × 8.315 × 298 J = – 11.15 kJ
 
Hope it helps
In case of any difficulty, please feel free to ask.
 
Regards
one month ago
Vikas TU
9163 Points
							
Combination of n- octane is 
C8H18 + 25/2 O2 -> 8CO2 +9H2O
del H = del E + RT deln 
del n = np - nr 
del n = 9+8 -25/2 = 4.5
del H - delE = 8.314 * 4.5 * 298 = 11.49 KJ 
Hope this helps 
Good Luck 
one month ago
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