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Question

The equilibrium concentration of $$B\left[ { \left( B \right)  }_{ e } \right]$$ for the reversible reaction $$A\rightleftharpoons B$$ can be evaluated by the expression:


A
Kc[A]1e
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B
KfKb[A]1e
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C
KfKb1[A]e
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D
KfKb[A]1
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Solution

The correct option is C $${ K }_{ f }{ { K }_{ b } }^{ -1 }{ \left[ A \right] }_{ e }$$
For given reaction, $$K_C= \cfrac {K_f}{K_b}=\cfrac {[B]_e}{[A]_e}$$
$$\therefore [B]_e= \cfrac {K_f}{K_b}[A]_e$$
$$= K_fK_b^{-1}[A]_e$$

Chemistry

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