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JEE Main 202431 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
31 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2024 · 31 January · Morning Shift · Chemistry · Chemical Equilibrium

For the given reaction, choose the correct expression of K _ C from...

For the given reaction, choose the correct expression of KC{K}_{C} from the following:

${Fe}^{3 +}\left(aq\right) + {SCN}^{-}\left(aq\right) ⇌ ( FeSCN{)}^{2 +}\left(aq\right)

Options

  1. A{K}_{C} = \dfrac{\left[\text{ FeSC }{N}^{2 +}\right]}{\left[F{e}^{3 +}\right]\left[SC{N}^{-}\right]}
  2. BKC=[Fe3+][SCN−][FeSCN2+]{K}_{C} = \dfrac{\left[F{e}^{3 +}\right]\left[SC{N}^{-}\right]}{\left[FeSC{N}^{2 +}\right]}
  3. C{K}_{C} = \dfrac{\left[\text{ FeSC }{N}^{2 +}\right]}{{\left[F{e}^{3 +}\right]}^{2}{\left[SC{N}^{-}\right]}^{2}}
  4. D{K}_{C} = \dfrac{{\left[\text{ FeSC }{N}^{2 +}\right]}^{2}}{\left[F{e}^{3 +}\right]\left[SC{N}^{-}\right]}

Correct answer

A. {K}_{C} = \dfrac{\left[\text{ FeSC }{N}^{2 +}\right]}{\left[F{e}^{3 +}\right]\left[SC{N}^{-}\right]}

Step-by-step solution

The equilibrium constant of a chemical reaction (usually denoted by the symbol K) provides insight into the relationship between the products and reactants when a chemical reaction reaches equilibrium. For example, the equilibrium constant of concentration (denoted by Kc) of a chemical reaction at equilibrium can be defined as the ratio of the concentration of products to the concentration of the reactants, each raised to their respective stoichiometric coefficients. It is important to note that there are different types of equilibrium constants that provide relationships between the products and the reactants of equilibrium reactions in terms of different units.

Equilibrium Constant Formula

Keqb=KfKb=[C]c[D]d[A]a[B]b=Kc{K}_{eqb} = \dfrac{{K}_{f}}{{K}_{b}} = \dfrac{{\left[C\right]}^{c}{\left[D\right]}^{d}}{{\left[A\right]}^{a}{\left[B\right]}^{b}} = {K}_{c}

Where Kc{K}_{c} indicates the equilibrium constant measured in moles per litre.

KC=[Products][Reactants]{K}_{C} = \dfrac{\left[Products\right]}{\left[Reactants\right]}

KC=[FeSCN2+][Fe3+][SCN−]{K}_{C} = \dfrac{\left[{FeSCN}^{2 +}\right]}{\left[{Fe}^{3 +}\right]\left[{SCN}^{-}\right]}

Hence, the answer is option A.

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