Question

Identify the equilibrium reaction in which $K_c$ has no units?

• A. Decomposition of $N{H}_3$
• B. Decomposition of $HI$
• C. Decomposition of $PC{l}_5$
• D. All of the above

Answer 1

The correct option is B Decomposition of $HI$

The units of Equilibrium constant $K_c$, will depend on the number of moles of reactants and products. When the total number of moles of products is equal to the total number of moles of reactants, then $K_c$ has no units.

In the question the reactions are expressed as:

Decomposition of $HI$
$2HI\left(g\right)\rightleftharpoons H_2\left(g\right)+I_2\left(g\right)$​


Decomposition of $PC{l}_5$
$PC{l}_5\left(g\right)\leftrightharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$

Decomposition of $N{H}_3$
$2N{H}_3\left(g\right)\leftrightharpoons N_2\left(g\right)+3H_2\left(g\right)$

In the above reactions, only in the reaction of decomposition of $HI$, moles of reactants and product are the same.

Hence Equilibrium constant $K_c$ has no units.


Theory:

Units of Equilibrium constant $K_c$:

The equilibrium constant for a general reaction,
$aA+bB\rightleftharpoons cC+dD$

$K_c=\frac{\left[C{]}_{eq}^c\right[D{]}_{eq}^d}{\left[A{]}_{eq}^a\right[B{]}_{eq}^b}$

where $\left[C{]}_{eq}^c\right[D{]}_{eq}^d\left[A{]}_{eq}^{a}and\right[B{]}_{eq}^b$ are the equilibrium concentrations of the reactants and products.

$K_c=\frac{\left(mol/L{)}^c\right(mol/L{)}^d}{\left(mol/L{)}^a\right(mol/L{)}^b}$

Unit of $K_c(mol/L{)}^{(c+d)-(a+b)}$