Is constant pressure heat capacity = entropy during adiabatic reversible expansion of perfect gas? For any process, ∆G = ∆H - ∆(TS).

Apply this on reversible adiabatic expansion of perfect gas, since the entropy of system does not change during this process, S is constant so

∆G = ∆H - S∆T

The process is reversible, so

∆G = 0 

and hence

∆H = S∆T

For perfect gas, assume heat capacity independent of temperature,

∆H = Cp∆T

and so

Cp∆T = S∆T.

∆T is not zero during adiabatic reversible expansion- temperature drops. So divide both sides by ∆T gives

Cp = S

Is this legit? If not where is my mistake? I have a hunch that I did something terribly stupid. The final result seems to be so counter-intuitive because

The Cp and S are constant throughout the process, so at the end of the process the two will still be the same. By measuring Cp, even though your system is not doing adiabatic reversible expansion, you will get S. So we are able to measure S?
Inspired by Chet Miller.
The fallacy in the argument is that, ΔG = 0  because the process is reversible. This is not true and so leading to wrong final result.
ΔG is a signpost for spontaneity of a process only if the process occurs through constant temperature and pressure. Likewise, that ΔG = 0 means reversibility, is true only if the process happens through constant temperature and constant pressure.
Adiabatic expansion clearly involves change in temperature, so ΔG is not a signpost for spontaneity of adiabatic change, and hence ΔG ≠ 0 for reversible adiabatic expansion.