Question

Predict expression from α in terms of $K_{eq}$ and concentration C :

$A_2 B_3(aq) \leftrightharpoons 2{A_3} (aq)+3B_{{2-}}(aq)$

• A. $\left(\frac{{K_{eq}}}{{108C^4}}\right)^{\frac{1}{5}}$
• B. $(\frac{K_{eq}}{5C^4})^\frac{1}{5}$
• C. $(\frac{4K_{eq}}{5C^4})^\frac{1}{5}$
• D. $(\frac{9K_{eq}}{108C^4})^\frac{1}{5}$

Answer 1

Answer: (A)

$\left(\frac{{K_{eq}}}{{108C^4}}\right)^{\frac{1}{5}}$

Answer 2

The expression in terms of $α$ (degree of dissociation), $K_{eq}$ (equilibrium constant), and concentration C for the given reaction $A_2 B_3(aq) \leftrightharpoons 2{A_3} (aq)+3B_{{2-}}(aq)$ can be derived as follows:

Let's assume the initial concentration of $A_2B_3$ is 'C'. At equilibrium, if α is the degree of dissociation, then the concentration of $A_{2}B_{3}$ will be $(1-α)$ C, and the concentrations of $A_3$ and $B_{2-}$ will be $2αC$ and $3αC$, respectively.

The equilibrium constant ($K_{eq}$) for the reaction is given by:

$K_{eq}$ = $\frac{([A_3]^2[B_{2-}]^3) }{ [A_2B_3]}$

Substituting the concentrations:

$K_{eq}$ =$\frac{ [(2αC)^2 \times (3αC)^3] }{ [(1 - α)C]}$

Simplifying:

$K_{eq} = \frac{(4α^2 \times 27α^3) }{ (1 - α)}$

$K_{eq} = \frac{(108α^5) }{ (1 - α)}$

To express this equation in terms of α, $K_{eq}$, and concentration C, we can rearrange it as:

$\left(\frac{{K_{eq}}}{{108C^4}}\right)^{\frac{1}{5}} = α$

Therefore, the expression in terms of α is:

$\left(\frac{{K_{eq}}}{{108C^4}}\right)^{\frac{1}{5}}$

Hence, the correct answer is option (A):$\left(\frac{{K_{eq}}}{{108C^4}}\right)^{\frac{1}{5}}$