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 ]

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 

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