At a given temperature, for a reversible reaction , if the concentration of reactants is doubled then the equilibrium costant will

(a) be doubled
(b) be halved
(c) change to 1/3
(d) remaine same

(d) remaine same

does it occur to you what constant in "equilibirum constant" means?

annapurna/ambika -

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constant
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To determine how the equilibrium constant will change when the concentration of reactants is doubled, we can refer to Le Chatelier's principle and the equilibrium expression for the reaction.

According to Le Chatelier's principle, when a system at equilibrium is disturbed by a change in concentration, pressure, or temperature, it will shift in a way that tends to counteract the change and restore equilibrium.

In the case of the given reversible reaction, if the concentration of reactants is doubled, it implies an increase in the concentration of the reactants. As a result, the system will attempt to counteract this increase by shifting the equilibrium position towards the products.

Now, let's examine the equilibrium expression for a general reversible reaction:

aA + bB ⇌ cC + dD

The equilibrium expression can be written as:

K = ([C]^c [D]^d) / ([A]^a [B]^b)

Where:
K = equilibrium constant
[A], [B], [C], [D] = concentrations of reactants and products at equilibrium
a, b, c, d = stoichiometric coefficients of the balanced equation

If we double the initial concentrations of the reactants ([A] and [B]), the new equilibrium concentrations will be 2[A] and 2[B]. Now let's substitute these new concentrations into the equilibrium expression:

K' = ([C]^c [D]^d) / ((2[A])^a (2[B])^b)

Simplifying further:

K' = ([C]^c [D]^d) / ([A]^a [B]^b * 2^a * 2^b)

Since 2^a * 2^b is equal to 2^(a + b), we can rewrite the expression as:

K' = ([C]^c [D]^d) / ([A]^a [B]^b * 2^(a + b))

Comparing the new equilibrium constant (K') with the original equilibrium constant (K), we can see that K' is equal to K divided by 2^(a + b). Thus, the equilibrium constant is halved when the concentration of reactants is doubled.

Therefore, the correct answer is (b) be halved.