Question

3.1 mol ${\rm{F}}{{\rm{e}}^{3 + }}$ and 3.2 mol of ${\rm{SC}}{{\rm{N}}^ - }$ are present in 1 L solution. At equilibrium 3 mol ${\rm{FeSC}}{{\rm{N}}^{2 + }}$ are formed. The equilibrium constant ${K_c}$ for the reaction $Fe^{+3}+SCN^{-}\rightleftharpoons FeSCN^{+2}$ will be:
(1) 150
(2) 400
(3) 300
(4) 250

Answer 1

We know that, the law of chemical equilibrium states that at a given temperature, a chemical system might reach a state in which a particular ratio of reactant and product concentrations has a constant value. This constant is termed as equilibrium constant $\left( {{K_c}} \right)$.

Complete step by step answer:
Here, first we have to make the ICE table for the reaction. The reaction is given as,

$Fe^{+3}+SCN^{-}\rightleftharpoons FeSCN^{+2}$

Given that initially the moles of ferric ion is 3.1 and moles of ${\rm{SC}}{{\rm{N}}^ - }$ is 3.2 mol. At equilibrium, the moles of the product are 3 mole. That means change in mole during the reaction is 3. So, the ICE table for the reaction is,

| ${\rm{F}}{{\rm{e}}^{3 + }}$| ${\rm{SC}}{{\rm{N}}^ - }$| ${\rm{FeSC}}{{\rm{N}}^{2 + }}$
---|---|---|---
Initial | 3.1 mol| 3.2 mol| 0
Change| -3 mol| -3 mol| +3 mol
Equilibrium | 3.1-3=0.1 mol| 3.2-3=0.2 mol| 3 mol

Now, we have to write the equilibrium constant expression for the reaction. Equilibrium constant is ratio of moles of product to reactant (gaseous).

${K_c} = \dfrac{3}{{0.1 \times 0.2}}$

$\Rightarrow {K_c} = \dfrac{3}{{0.02}} = 150$

Therefore, the equilibrium constant for the reaction is 150.

So, the correct answer is Option 1.

Additional Information:
Let’s learn about equilibrium constant in detail. Equilibrium constant helps in quantitative measurement of strength of bases and acids in water.

${K_a}$ is acid constant for acidic reaction.

$HA+H_2O\rightleftharpoons H_3O^{+}+A^{-}$

For this reaction, acidic constant is,

${K_a} = \dfrac{{\left[ {{{\rm{H}}_{\rm{3}}}{{\rm{O}}^ + }} \right]\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}}$

${K_b}$ is the basic constant for equilibrium reaction.

$BOH+H_2O\rightleftharpoons B^{+}+OH^{-}$

${K_b} = \dfrac{{\left[ {{{\rm{B}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^ - }} \right]}}{{\left[ {{\rm{BOH}}} \right]}}$

Note: Always remember that, in equilibrium constant expression only gaseous reactants and products are considered. The value of equilibrium constant is dependent on temperature. Its value is independent of the actual amount of product and reactant, presence of catalyst and presence of inert material, pressure and volume of reactants and products.