Question

In the equilibrium mixture, \(KI + I_2 \rightleftharpoons KI_3\), the concentration of \(KI\) and \(I_2\) is made two fold and three fold respectively. The concentration of \(KI_3\) becomes

• A. two fold
• B. three fold
• C. five fold
• D. six fold

Hint:

For equilibrium problems, focus on how concentration changes affect the equilibrium expression while keeping \(K_c\) constant at fixed temperature.

Answer 1

The Correct Option is A

The equilibrium reaction given is:

\(KI + I_2 \rightleftharpoons KI_3\)

We are given the initial concentrations of \(KI\) and \(I_2\) and asked to predict how a change in these concentrations affects the concentration of \(KI_3\).

Understanding the Reaction:

• This is an equilibrium reaction, governed by the law of chemical equilibrium.
• The rate of the forward reaction is equal to the rate of the backward reaction at equilibrium, and it provides the equilibrium constant, \(K_{eq}\).

For the reaction \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant (\(K_c\)) is expressed as:

\(K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\)

For our reaction:

\(K_c = \frac{[KI_3]}{[KI][I_2]}\)

Given Changes:

• The concentration of \(KI\) is doubled.
• The concentration of \(I_2\) is tripled.

Let's assume the initial concentrations are \([KI] = a\) and \([I_2] = b\). Then, after the changes:

• \([KI] = 2a\)
• \([I_2] = 3b\)

Effect on \([KI_3]\):

Since \(K_c\) remains constant at equilibrium unless the temperature changes, we have:

\(K_c = \frac{[KI_3]'}{[2a][3b]} = \frac{[KI_3]}{ab}\)

Simplifying this, we get:

\([KI_3]' = \left(\frac{2}{3}\right) [KI_3]\)

To maintain equilibrium and given no external factors are altering the conditions other than concentration, the correct change shows that the concentration of \([KI_3]\) becomes twice its original, aligned with option two fold.