Question

`(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` ?
A. `-40KJ//mol`
B. `-50KJ//mol`
C. `-45KJ//mol`
D. `-6KJ//mol`

Answer 1

Correct Answer - B
`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`