For a reversible reaction if the concentrations of the reactants are doubled then the equilibrium:

be halved

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be doubled

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remain the same

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become

1 / 4^{t h}

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Solution

The correct option is C remain the same
For a given reaction , at equilibrium $K_{c} = \frac{K_{f}}{K_{r}} = \frac{c o n c e n t r a t i o n o f p r o d u c t s}{c o n c e n t r a t i o n o f r e a c t a n t s}$

Therefore $K_{c}$ is independent of change in concentration.

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A reversible chemical reaction having two reactants in equilibrium. If the

concentrations of the reactants are doubled, then the equilibrium constant

will

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