Question

The equilibrium ∆G° (HI;g) ≅ +1.7 kJ. What is the

equilibrium constant at 250
°C for
2HI(g) ⇌ H2 (g) + I2 (g)


(a) 24.0 (b) 3.9

(c) 2.0 (d) 0.5

Answer 1

Hola User______________

Here is Your Answer...!!!
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⭐Actually welcome to the concept of the GIBBS FREE ENERGY..

⭐basically we know that. ..

⭐at equilibrium we get the equation as ..


⭐Delta G° = - RT (2.303) log ( Keq)

⭐so we get as...

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Answer 2

Answer:

given 2HI⇌H−2(g)+I2(g)   eq. constant K at 25oC

ΔG=1.7 kJ

we know   ΔG=RTln K

1.7×103=2.303logK(RT)

1700=2.303(8.314)(298)logK

logK=2.303×8.314×2981700=57061700

Hence logK = 0.29

K=1.94≈2