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Question

In a chemical equilibrium $$A+B\rightleftharpoons C+D$$ when one mole each of the two reactants are mixed, $$0.4\ mole$$ each of the products are formed.The equilibrium constant calculated is :-


A
1
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B
0.36
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C
2.25
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D
49
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Solution

The correct option is D $$\dfrac{4}{9}$$
$$\underset {1\\(1-x)}{A}\ +\underset {1\\(1-x)}{B}\rightleftharpoons \underset {0\\x}{C}+\ \underset{0\\x}{D}$$


$$\displaystyle K=\frac {[C][D]}{[A][B]}=\frac {x\times x}{(1-x)(1-x)}$$

A/Q
$$x=0.4$$

Hence, $$\displaystyle K=\frac {0.4^2}{0.6^2}=\frac 49$$

Hence, option D is correct

Chemistry

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