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Calculate the entropy change in system if 2 mole of methane undergoes complete combustion at 300 K from the following data. Given data `:DeltaH_("cumbustion")^(@) CH_(4)(g)=-900kJ,DeltaG_(f)^(@)CH_(4)(g)=-40,DeltaG_(f)^(@)(H_(2)O)(l)=-120, DeltaG_(f)^(@)CO_(2)(g)=-400kJ//mol` Instruction:- If your answer is -ve, then wire double the magnitude as your final answer For example : If `DeltaS_("surr")=+20J/K` then write your answer as 20 but if `DeltaS_("surr")=-20` then write your answer as 40 Express your answer in J/K. |
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Answer» [4000] `2CH_(4) to 4O_(2) to 2CO_(2)++4H_(2)O` `DeltaH^(@)=-900xx2=-1800kJ` `DeltaG^(@)=DeltaG_("f Products")^(@)-DeltaG_("f Reactant")^(@)` `=(-400)xx2+4(-120)-[2xx(-40)+0]` =-800-480+80 =-1200kJ/mol `DeltaG^(@)=DeltH^(@)-TDeltaS^(@)` `DeltaS_("system")^(@)=(DeltaH^(@)-DeltaG^(@))/(T)` `((-1800)-(-1200)kJ)/(300)=(-600)/(300)kJ` `-2kJ=-2000J//K` Final answer4000 |
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