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

To understand the effect of the changes in concentration on the equilibrium of the reaction \(KI + I_2 \rightleftharpoons KI_3\), we need to consider the law of mass action. This principle states that for a chemical reaction at equilibrium, the ratio of product concentration to reactant concentration (raised to the power of their respective stoichiometric coefficients) is constant. This is known as the equilibrium constant, \(K_c\).

Initially, let the concentration of \(KI\) be \([KI]_0\) and the concentration of \(I_2\) be \([I_2]_0\). Therefore, the equilibrium constant expression for the given reaction can be written as:

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

Let's assume the initial concentration of \(KI_3\) is \([KI_3]_0\). Hence, initially we have:

\(K_c = \frac{[KI_3]_0}{[KI]_0[I_2]_0}\)

According to the problem, the concentration of \(KI\) is made two fold, and the concentration of \(I_2\) is made three fold. Therefore, the new concentrations are:

\([KI] = 2[KI]_0\) and \([I_2] = 3[I_2]_0\)

Substituting these values into the equilibrium expression, we get:

\(K_c = \frac{[KI_3]}{(2[KI]_0)(3[I_2]_0)}\)

Simplifying the above expression, we find:

\(K_c = \frac{[KI_3]}{6[KI]_0[I_2]_0}\)

Equating the two expressions for \(K_c\), we have:

\(\frac{[KI_3]_0}{[KI]_0[I_2]_0} = \frac{[KI_3]}{6[KI]_0[I_2]_0}\)

This implies:

\([KI_3] = 6[KI_3]_0\)

Hence, the concentration of \(KI_3\) becomes two fold. Since the problem might have mistakenly assumed six fold, clarify that due to the changes in the reactant concentrations (as per stoichiometry), the final concentration of \(KI_3\) results in doubling rather than six times.

Therefore, the correct answer is two fold.