the K_p value for the equilibrium H₂ (g) +I₂ (s) ⇌ 2HI (g) is 871 at 25 deg . if the vapour pressure of iodine is 4 x 10^-4 atm, calculate the equilibrium constant in term of partial pressures at the same temperature for the reaction ; H₂ (g) + I₂ (s) ⇌ 2HI (g).

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Solution

Dear Suruchi Sharma,

$G i v e n; I_{2} (s) \Leftrightarrow I_{2} (g) - - - - 1 v a p o u r p r e s s u r e o f I_{2} (g) = 4 \times 10^{- 10} a t m H_{2} (g) + I_{2} (s) \Leftrightarrow 2 H I (g) - - - - 2 {K_{p}}_{2} f o r e q 2 = \frac{(p_{H I}) 2}{p_{H_{2}}} = 871 f r o m e q 1 a n d 2, w e g e t H_{2} (g) + I_{2} (g) \Leftrightarrow 2 H I (g) - - - - 3 K_{p 3} f o r e q 3 = \frac{(p_{H I}) 2}{[p_{H_{2}}] [[p_{I_{2}}]} p u t t h e v a l u e o f {K_{p}}_{2} a n d v a p o u r p r e s s u r e o f I_{2} (g) = 4 \times 10^{- 10} a t m K_{p 3} = \frac{871}{4 \times 10^{- 10}} = 2.18 \times 10^{12} S o t h e e q u i l i b r i u m c o n s t a n t w . r . t . p a r t i a l p r e s s u r e i s 2.18 \times 10^{12}$

Regards

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