For a system, `a+2BhArrC` , the equilibrium comcentrations are `[A]=0.06, [B]=0.12`, and `[C]=0.216`. The `K_(c )` for the reaction isA. `120`B. `400`C. `4xx10^(-3)`D. `250`

For a system, `a+2BhArrC` , the equilibrium comcentrations are `[A]=0.

1.	For a system, `a+2BhArrC` , the equilibrium comcentrations are `[A]=0.06, [B]=0.12`, and `[C]=0.216`. The `K_(c )` for the reaction isA. `120`B. `400`C. `4xx10^(-3)`D. `250`
Answer» Correct Answer - D `A+2BhArrC` `K=([C])/([A][B]^(2))rArr ((0.216))/((0.06)(0.012)^(2))=250`

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For a system, `a+2BhArrC` , the equilibrium comcentrations are `[A]=0.06, [B]=0.12`, and `[C]=0.216`. The `K_(c )` for the reaction isA. `120`B. `400`C. `4xx10^(-3)`D. `250`

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