`DeltaH` for the reaction, `C("graphite")+2H_(2)(g) to CH_(4)(g)` at 298 K and 1 atm is -17900 cal. The `DeltaE` for the above conversion should be
`DeltaH` for the reaction,
`C("graphite")+2H_(2)(g) to CH_(4)(g)`
at 298 K and 1 atm is -17900 cal. The `DeltaE` for the above conversion should be
A. `-17900` cal
B. 17900 cal
C. 17304 cal
D. `-17304` cal
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Correct Answer - D
`C("graphite")+2H_(2)(g)toCH_(4)(g),DeltaH=-17900` cal
We have, `DeltaH=DeltaE+Deltan_(g)RT`
`therefore DeltaH=DeltaE-1xxRxxT`
`-17900=DeltaE-1xx298xx2`
`DeltaE=-17900+596=-17304cal`
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