For the reaction, `A+2B hArr C`, the expession for equilibrium constant isA. `([A][B]^(2))/([C])`B. `([A][B])/([C])`C. `([C])/([A][B]^(2))`D. `([C])/([2B][A])`

For the reaction, `A+2B hArr C`, the expession for equilibrium constan

1.	For the reaction, `A+2B hArr C`, the expession for equilibrium constant isA. `([A][B]^(2))/([C])`B. `([A][B])/([C])`C. `([C])/([A][B]^(2))`D. `([C])/([2B][A])`
Answer» Correct Answer - C `A+2B overset(K)(hArr) C` According to law of mass action `K=([C])/([A][B]^(2))`

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For the reaction, `A+2B hArr C`, the expession for equilibrium constant isA. `([A][B]^(2))/([C])`B. `([A][B])/([C])`C. `([C])/([A][B]^(2))`D. `([C])/([2B][A])`

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