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Reaction of methanol with dioxygen was c...

Reaction of methanol with dioxygen was carried out and `DeltaU` was found to be `-726"kJ mol"^(-1)` at 298 K. The enthalpy change for the reaction will be
`CH_(3)OH_((l))+(3)/(2)O_(2(g)) rarr CO_(2(g))+2H_(2)O_((l)) , DeltaH=-726"kJ mol"^(-1)`

A

`-741.5" kJ mol"^(-1)`

B

`-724.7" kJ mol"^(-1)`

C

`+741.5" kJ mol"^(-1)`

D

`+727.2" kJ mol"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B
`Delta n_(g) =1-(3)/(2) =-(1)/(2)` ltbr `Delta H = Delta U + n_(g)RT`
`=-726xx1000" J mol"^(-1)+(-(1)/(2)xx8.314J K^(-1)" mol"^(-1)xx298 K)`
`=-727238.78" J mol"^(-1)=-727" kJ mol"^(-1)`
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