For the given reaction, choose the correct expression of $\mathrm{K}_{\mathrm{C}}$ from the following :- $\mathrm{Fe}_{(\mathrm{aq})}^{3+}+\mathrm{SCN}_{(\mathrm{aq})}^{-} \rightleftharpoons(\mathrm{FeSCN})_{(\mathrm{aq})}^{2+}$

(A) $\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}$

(B) $\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}{\left[\mathrm{FeSCN}^{2+}\right]}$

(C) $\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]^{2}\left[\mathrm{SCN}^{-}\right]^{2}}$

(D) $\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]^{2}}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}$

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Equilibrium

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For the given reaction, choose the correct expression of

K_{C}

from the following :-

Fe_{(aq)}^{3 +} + SCN_{(aq)}^{-} ⇌ (FeSCN)_{(aq)}^{2 +}

(A) $K_{C} = \frac{[ FeSCN ^{2 +} ]}{[ Fe ^{3 +} ] [ SCN ^{-} ]}$ ✓ Correct
(B) $K_{C} = \frac{[ Fe ^{3 +} ] [ SCN ^{-} ]}{[ FeSCN ^{2 +} ]}$
(C) $K_{C} = \frac{[ FeSCN ^{2 +} ]}{[ Fe ^{3 +} ] ^{2} [ SCN ^{-} ] ^{2}}$
(D) $K_{C} = \frac{[ FeSCN ^{2 +} ] ^{2}}{[ Fe ^{3 +} ] [ SCN ^{-} ]}$

Answer: (A)

Solution

For the equilibrium

F e_{(aq)}^{3 +} + SC N_{(aq)}^{-} ⇌ [FeSCN]_{(aq)}^{2 +}

, the law of mass action gives the equilibrium constant as products over reactants, each raised to its stoichiometric coefficient (all equal to 1 here):

K_{C} = \frac{[ FeSC N ^{2 +} ]}{[ F e ^{3 +} ] [ SC N ^{-} ]}

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option

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JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Equilibrium

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