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At 500 K, equilibrium constant, `K_(c)`, for the following reaction is 5.
`(1)/(2) H_(2) (g) + (1)/(2) I_(2) (g) hArr HI(g) `
What would be the equilibrium constant `K_(c)` for the reaction,
`2 HI (g) hArr H_(2) (g) +I_(2) (g) `?
A. 0.04
B. 0.4
C. 25
D. 2.5

1 Answer

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Best answer
Correct Answer - A
For `(1)/(2) H_(2) + (1)/(2) I_(2) hArr HI, K = 5 `.
`:. ` For `H_(2) + I_(2) hArr 2 HI, K = 5^(2) = 25`
For reverse reaction, `2HI hArr H_(2)+I_(2). K = (1)/(25) = 0.04`.

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