The equilibrium constant Kp for the reaction : H2g+I2g⇐2HIg is 130 at 510 K Calculate ∆G for the - Chemistry - Electrochemistry

Question

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 ]

Vartika Jain · Accepted Answer

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

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 ]

The equilibrium constant K_p for the reaction : $H_{2 (g)} + I_{2 (g)} \Leftarrow 2 H I_{(g)}$ is 130 at 510 K. Calculate $∆ G °$ for the following reaction at the same temperature :
$2 H I_{(g)} \Leftarrow H_{2 (g)} + I_{2 (g)}$ [Given : R = 8.314 J K^-1 mol^-1 ]