Prove that for an ideal-gas reaction, #(dlnK_C^@)/(dT) = (DeltaU^@)/(RT^2)#?
I found this question particularly difficult, but I figured it out and wanted to share.
I found this question particularly difficult, but I figured it out and wanted to share.
1 Answer
For an ideal gas reaction, we begin with the definition of
#K_C^@ = K_P^@((P^@)/(RTc^@))^(Deltan)#
and differentiate
#(dlnK_C^@)/(dT)#
#= d/(dT)[ln{e^(-(DeltaG^@)/(RT))((P^@)/(RTc^@))^(Deltan)}]#
#= d/(dT)[-(DeltaG^@)/(RT) + Deltanln((P^@)/(RTc^@))]#
#= d/(dT)[-(DeltaG^@)/(RT)] + Deltan(cancel(RTc^@)/cancel(P^@))*-cancel(P^@)/(cancel(Rc^@)T^cancel(2))#
#= stackrel("Product Rule")overbrace((DeltaG^@)/(RT^2) - 1/(RT)(dDeltaG^@)/(dT)) - (Deltan)/T#
By definition, since
#dG^@ = -S^@dT + VdP^@#
or
#dDeltaG^@ = -DeltaS^@dT + DeltaVdP^@#
and acquire
#((delDeltaG^@)/(delT))_(P^@) = (dDeltaG^@)/(dT) = -DeltaS^@#
Then we proceed to acquire the result by noting that
#=> (DeltaG^@)/(RT^2) + (DeltaS^@)/(RT) - (Deltan)/T = (DeltaG^@)/(RT^2) + (TDeltaS^@)/(RT^2) - (DeltanRT)/(RT^2)#
#= (DeltaG^@ + TDeltaS^@ - DeltanRT)/(RT^2) = (DeltaH^@ - cancel(TDeltaS^@ + TDeltaS^@) - DeltanRT)/(RT^2)#
#= (DeltaU^@ + Delta(PV) - DeltanRT)/(RT^2) = (DeltaU^@ + cancel(DeltanRT) - cancel(DeltanRT))/(RT^2)#
#= color(blue)((DeltaU^@)/(RT^2))#