The equilibrium constant Kp for the reaction : H 2 g + I 2 g ⇐ 2 H I g is 130 at 510 K. Calculate ∆ G ° for the following reaction at the same temperature : 2 H I g ⇐ H 2 g + I 2 g [Given : R = 8.314 J K-1 mol-1 ] Share with your friends Share 1 Vartika Jain answered this Dear Student, For the reaction, H2(g) + I2(g) ↔ 2HI(g)Kp= (pHI)2(pH2)(pI2)=130Now, for the reaction2HI(g) ↔ H2(g) + I2(g)Kp'=(pH2)(pI2)(pHI)2i.e. Kp'=1Kpor, Kp'=1130=7.69×10-3In the reaction, ∆n = 2-2=0and Kp=Kc(RT)∆n=Kc(RT)0=Kcor, Kp=Kc=7.69×10-3and, ∆G°=2.303RTlogKC =2.303×8.314×510×log (7.69×10-3) =-20,644.01 Jmol-1 =-20.644 kJmol-1 2 View Full Answer