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the value of equilibrium constant for the reaction `HI(g) hArr 1//2 H_(2) (g) +1//2 I_(2) (g) " is " 8.0` The equilibrium constant for the reaction `H_(2) (g) +I_(2) (g) hArr 2HI(g)` will beA. 16B. `1//8`C. `1//16`D. `1//64` |
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Answer» Correct Answer - D `HI(g) hArr 1//2 H_(2) (g) + 1//2I_(2) (g)` `K= [[H_(2)]^(1//2) [I_(2)]^ (1//2)]/[[HI]] =8` `H_(2)(g) +I_(2) (g) hArr 2HI(g)` `K =[[H]^(2)]/[[H_(2)][I_(2)]] =((1)/(8))^(2) =(1)/(64)` |
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