Question

A certain isothermal and isobaric process is characterized by DG = + 150. kJ at 500. K and and DG = + 350. kJ at 600. K. Assuming DH and DS do not depend on temperature, determine the temperature where the process occurs reversibly and the temperature range where it occurs spontaneously.

Answer 1

For a process to be reversible:

dG < 0, i.e. negative

so...

dG = dH - T*dS

dG = 150 kJ at T = 500 K

dG = 350 kJ at T = 600 K

then:

dG1 = dH - T1dS

dG2 = dH - T2*dS

substituite data:

150 = dH - 500*dS

350= dH - 600*dS

solve for dS:

(350-150) = -100dS

dS = 200/-100 = -2 kJ

then

150 = dH - 500*(-2)

dH = 150-1000 = -850 kJ

so...

for dG < 0

dH - T*dS < 0

dH < T*dS

-850 < T*(-2)

-850/(-2) > T

T < 425 K

as long as T is less than 425, this process will be spontanous