JEE Main 2024 · 31 January · morning · Single correct · +4 / −1
For the given reaction, choose the correct expression of K_C from the following :-
Fe_((aq))³⁺+SCN_((aq))⁻⇌ (FeSCN)_((aq))²⁺
- (a) K_C=[Fe³⁺][SCN⁻]/[FeSCN²⁺]
- (b) K_C=[FeSCN²⁺]/[Fe³⁺][SCN⁻]
- (c) K_C=[FeSCN²⁺]²/[Fe³⁺][SCN⁻]
- (d) K_C=[FeSCN²⁺]/[Fe³⁺]²[SCN⁻]²
Answer and solution
Answer (b) K_C=[FeSCN²⁺]/[Fe³⁺][SCN⁻]
Solution
Chemistry ma’am · 3 steps
Products on top, reactants below, and each concentration raised to its own coefficient.
- Law of mass action. For aA + bB ⇌ cC, the equilibrium constant is K_c = [C]ᶜ/[A]ᵃ[B]ᵇ. The power on each concentration is that species' coefficient in the balanced equation.
- Read the coefficients. In Fe³⁺ + SCN⁻ ⇌ FeSCN²⁺, every species has a coefficient of 1. So every concentration is raised to the power 1, with no squares anywhere.
K_c = [FeSCN²⁺]/[Fe³⁺][SCN⁻]
- So. The product FeSCN²⁺ goes in the numerator and both reactant ions go in the denominator, each to the first power. That is option (b).
The idea to keep. Write K_c straight from the balanced equation: products over reactants, and each power equals the stoichiometric coefficient. Check the coefficients before you put in any squares.
Answer from the NTA final answer key. Solution by PrepWiser.