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Identificati formula moleculara a alchinei (CnH2n-2) care prezinta entalpia de formare delta f(indice) H = +226,75kj/mol si care la arderea a 44,64 moli alchina, rezulta 56052,2 kj. Se cunosc entalpiile de formare standard:
delta f(indice) Hco2= - 393kj/mol si delta f(indice) Hh2o = -285,5 kj/mol.


Răspuns :

CnH(2n-2) + (3n-1)/2 O2 ⇒ nCO2 +(n-1)H2O    ΔH = 56052,2/44,64 = -1255,65
ΔH = nΔH CO2 + (n-1)ΔH H2O - ΔH CnH(2n-2)
-1255,65 = - 393n - 285,5(n-1) - 226,75
678,5n = 1314,4    n = 2
C2H2