Are enthalpy and entropy affected by temperature?

Question

Mateo Aguilar · Accepted Answer

They are state functions of temperature, yes.

#color(blue)(DeltaH(T) = int_(T_1)^(T_2) C_P dT) = C_P(T_2 - T_1)#

#color(blue)(DeltaS(T) = int_(T_1)^(T_2) C_P/T dT) = C_Pln|T_2/T_1|#

If we assume that the constant-pressure heat capacity #C_P# is constant (which is a pretty decent assumption for small temperature ranges), then we can pull it out of the integral and focus on integrating the temperature.

(Normally though, #C_P = C_P(T)#.)

This indicates that entropy and enthalpy vary in direct proportion to temperature.

Below is the derivation of this.

The constant-pressure heat capacity is defined as the change in enthalpy with respect to temperature at a constant pressure.

#\mathbf(((delH)/(delT))_P = (delq_p)/(delT) = C_P)#

where #q_p# is heat flow at a constant pressure.

This allows us to write under constant pressure:

#dH = delq_p = C_PdT#

Now combine all of them to obtain:

#color(green)(DeltaH = q_p = int_(T_1)^(T_2) C_PdT) = C_P(T_2 - T_1)#

Because of that relationship between #q_p# and #DeltaH#, notice how we can relate two similar equations, provided the heat flow is reversible:

#DeltaH = q_(p,rev)#

#color(green)(DeltaS = q_(p,rev)/T)#

With the green relationships mentioned above, we obtain:

#color(blue)(DeltaS = int_(T_1)^(T_2) C_P/T dT)#

Lastly, we highlight the fact that entropy and enthalpy are both functions of temperature by writing:

#DeltaH = DeltaH(T)#

#DeltaS = DeltaS(T)#

Peyton Adams · Answer

undefined Yes, enthalpy and entropy are affected by temperature. Entropy generally increases with temperature, while enthalpy change depends on the specific reaction or process.