Question

using the Gibbs -Helmholtz equation Consider the equilibrium of ethane gas with the decomposition products hydrogen gas and ethylene gas at 1000 K. The value of Kp at 1000 K is 1.41 x 10-1 and H 0 =137 kJmol-1 . Using this information, find Kp at 298. Assume H 0 is constant with temperature. Does your result match what you expect according to Le Chatelier’s principle?

Answer 1

C2H6----------->C2H4+H2

Kp= [C2H4][H2]/[C2H6]

Kp = 1.41/10=0.141 and deltaH= 137 Kj/mole=137000 J/mole

since deltaG= -RTlnKp= -8.314*1000*ln(0.141)=16287 J/mole

since deltaG= deltaH-T*deltaS

deltaS = (deltaH-deltaG)/T= (137000-16287)/1000=120 J/Mole.K

at 298K, deltaG= deltaH-T*deltaS becomes 137000-298*120 =101240 J/mole

-RTlnK= 101240

K at 298K= exp^{{(-101240/(298*8.314)}}=1.79*10^-18

The reaction is endothemic so decrease in the temperature favors exothermic direction. So more C2H6 is formed.

so decreased temperature takes reaction backward and more C2H6 is formed which reduces the Kp value.