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$$4$$ moles of $$A$$ are mixed with $$4$$ moles of $$B$$ when $$2$$ moles of $$C$$ are formed at equilibrium according to the reaction $$A+B\rightleftharpoons C+D$$.
The value of the equilibrium constant is :


A
4
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B
1
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C
1/2
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D
1/4
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Solution

The correct option is B $$1$$
$$\underset {4\\4-x}{A}+\underset {4\\4-x}{B}\rightleftharpoons \underset {0\\x}{C} +\underset {0\\x}{D}$$
$$\therefore [A]=4-2=2$$   
    $$[B]=4-2=2$$
    $$[C]=2,   [D]=2$$

The value of the equilibrium constant is, $$K=\dfrac {[C][D]}{[A][B]}=\dfrac {2\times 2}{2\times 2}=1$$

The correct option is $$B$$

Chemistry

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