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For the system `3A+2BhArrC,` the expression for equilibrium constant isA. `([3A][2B])/(C)`B. `([C])/([3A][2B])`C. `([A]^(2)[B]^(2))/([C])`D. `([C])/([3A]^(3)[2B]^(2))` |
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Answer» Correct Answer - D Equlibrium constant for the reaction, `3A+2BhArrC isK=([C])/([A]^(3)[B]^(2)).` |
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