Question

b) i) At 300°C, the equilibrium constant for the reaction H2 + 12 = 2HI
is 90. Calculate the change in free energy as for this reaction.​

Answer 1

Answer:

-21394.052 J

Explanation:

Given K= 90

T= 300°C = 300+273= 573 K

R = 8.314 J mol^-1 K^-1

∆G= -2.303 RT log K

= -2.303 * 8.314*573*log 90

= -21394.052 J