InterviewSolution
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InterviewSolution
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Given `S^@` for `C_("Graphite"), H_(2(g))` and `CH_(4(g))` are `5.70, 130.7` and `186.3 JK^(-1) mol^(-1)`. Also standard heat of formation of `CH_4` is `-74.81kJmol^(-1).` Calculate the standard Gibbs energy change for the formation of methane at `298K.` `C_("Graphite")+2H_(2(g))toCH_(4(g))` |
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Answer» `Delta_(f)H^(@)CH_(4)=H_(CH_(4))^(@)=-74.81kJmol^(-1)` Also `H^(@)` for `C_("graphite")` and `H_(2(g))=0` (standard state of elements) `C_("graphite")+2H_(2(g))toCH_(4(g)),` `:. DeltaH^(@)=DeltaH_(CH_(4))^(@)-H_(C_("graphite"))^(@)-2xxH_(H_(2))^(@)` `=74.81` Also `DeltaS^(@)` (for the reaction) `=S_(CH_(4))^(@)-S_(C_("graphite"))^(@)-2xxS_(H_(2))^(@)` `=186.3-5.70-130.7xx2=-80.8JK^(-1)mol^(-1)` Using `DeltaG^(@)=DeltaH^(@)-TDeltaS^(@)` `=-74.81-298xx[-80.8xx10^(-3)]` `=-50.73 kJ` |
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