1.

In the following equilibria `I:A+2B hArr C, K_(eq)=K_(1)` `II: C+DhArr 3A, K_(eq)=K_(2)` `III, 6B+D hArr 2C, K_(eq)=K_(3)` Hence,A. `3K_(1)+K_(2)=K_(3)`B. `K_(1)^(3).K_(2)^(2)=K_(3)`C. `3K_(1)+K_(2)^(2)=K_(3)`D. `K_(1)^(3).K_(2)=K_(3)`

Answer» Correct Answer - D
To obtain `K_(3)` multiply eqn. I by 3 and add eqn. II
`3A+6B hArr 3C, K_(eq)=K_(1)^(3)`
`C+D hArr 3A , K_(eq)=K_(2)`
Adding, we get
`6B+D hArr 2C, K^(eq)=K_(3)=K_(1)^(3).K_(2)`


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