(1)/(2)N_(2)(g)+O_(2)rarrNO_(2)(g),Delta_(r)H^(@)=-40KJ//mol Given: C_(P.m)(NO_(2),g)=40J//mol//K,C_(p,m) (O_(2),g)=30JK^(-1)mol^(-1) C_(P,m)N_(2)(g)=30JK^(-1)mol^(-1) What is the enthalpy of formation of NO_(2)(g) at 1298K ?

(1)/(2)N_(2)(g)+O_(2)rarrNO_(2)(g),Delta_(r)H^(@)=-40KJ//mol Given: C

1.	(1)/(2)N_(2)(g)+O_(2)rarrNO_(2)(g),Delta_(r)H^(@)=-40KJ//mol Given: C_(P.m)(NO_(2),g)=40J//mol//K,C_(p,m) (O_(2),g)=30JK^(-1)mol^(-1) C_(P,m)N_(2)(g)=30JK^(-1)mol^(-1) What is the enthalpy of formation of NO_(2)(g) at 1298K ?
Answer» <P>`-40KJ//mol` `-50KJ//mol` `-45KJ//mol` `-6KJ//mol` SOLUTION :`Delta_(r)H_(T)=Delta_(r)H^(@)+int_(298)^(1298)Delta_(r)C_(P)dT` At `1298KDelta_(r)H=-40KJ-5` `DeltaT=-40KJ-5xx1000xx10^(-3)KJ=-45KJ//mol`