A) \[+205\text{ }kJ\text{ }mo{{l}^{-1}}\]
B) \[-205\text{ }kJ\text{ }mo{{l}^{-1}}\]
C) \[+605\text{ }kJ\text{ }mo{{l}^{-1}}\]
D) \[-605\text{ }kJ\text{ }mo{{l}^{-1}}\]
Correct Answer: D
Solution :
\[3{{C}_{2}}{{H}_{2}}\xrightarrow{Polymerisation}{{C}_{6}}{{H}_{6}}\] \[\Delta H\] = enthalpy of product - enthalpy of reactant \[=85-3(230)\] \[=85-690\] \[=-605\,kJ\,mo{{l}^{-1}}\]You need to login to perform this action.
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