Prove that for an ideal-gas reaction, #(dlnK_C^@)/(dT) = (DeltaU^@)/(RT^2)#?

Question

I found this question particularly difficult, but I figured it out and wanted to share.

Henry Antrim · Accepted Answer

For an ideal gas reaction, we begin with the definition of #K_C^@#

#K_C^@ = K_P^@((P^@)/(RTc^@))^(Deltan)#

and differentiate #lnK_C^@# with respect to #T#. We would obtain, noting the usage of the product rule on #K_P^@ = e^(-DeltaG^@/(RT))# (since #DeltaG^@ = DeltaG^@(T)#):

#(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 #DeltaG^@# is defined for a fixed #"1 bar"# pressure, we can refer to the Maxwell relation

#dG^@ = -S^@dT + VdP^@#

or

#dDeltaG^@ = -DeltaS^@dT + DeltaVdP^@#

and obtain

#((delDeltaG^@)/(delT))_(P^@) = (dDeltaG^@)/(dT) = -DeltaS^@#

Then we proceed to acquire the result by noting that #DeltaG^@ = DeltaH^@ - TDeltaS^@#, and that for an ideal gas, #DeltaH^@ = DeltaU^@ + Delta(PV) = DeltaU^@ + DeltanRT#:

#=> (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))#

Cooper Anderson · Answer

undefined For an ideal-gas reaction, ΔU^@ = ΔH^@. Using the Van't Hoff equation, ln(K_C^@2/K_C^@1) = -(ΔH^@/R)(1/T^2_2 - 1/T^2_1). Take the derivative of both sides with respect to T and solve for (dlnK_C^@)/(dT) to get (DeltaU^@)/(RT^2).