For the following reaction, the equilibrium constant \(\mathrm{K}_{\mathrm{c}}\) at 298 K is \(1.6 \times 10^{17}\). \[ \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{S}^{2-}(\mathrm{aq}) \rightleftharpoons \mathrm{FeS}(\mathrm{~s}) \] When equal volumes of \(0.06 \mathrm{M} \mathrm{Fe}^{2+}(\mathrm{aq})\) and \(0.2 \mathrm{M} \mathrm{S}^{2-}(\mathrm{aq})\) solutions are mixed, the equilibrium concentration of \(\mathrm{Fe}^{2+}(\mathrm{aq})\) is found to be \(\mathbf{Y} \times 10^{-17} \mathbf{M}\). The value of Y is ____

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Equilibrium

For the following reaction, the equilibrium constant

K_{c}

at 298 K is

1.6 \times 1 0^{17}

Fe^{2 +} (aq) + S^{2 -} (aq) ⇌ FeS (s)

When equal volumes of

0.06 M Fe^{2 +} (aq)

and

0.2 M S^{2 -} (aq)

solutions are mixed, the equilibrium concentration of

Fe^{2 +} (aq)

is found to be

Y \times 1 0^{- 17} M

. The value of Y is ____

Answer: 8.9

When equal volumes are mixed, initial concentrations are halved:

[Fe^{2 +}]_{0} = 0.03 M

[S^{2 -}]_{0} = 0.1 M

.

Let

x

mol/L of

Fe^{2 +}

react. At equilibrium:

[Fe^{2 +}] = 0.03 - x

[S^{2 -}] = 0.1 - x

.

Since

K_{c} = 1.6 \times 1 0^{17}

is very large, the reaction goes nearly to completion. The limiting reagent is

Fe^{2 +}

, so

x \approx 0.03

.

At equilibrium:

[Fe^{2 +}] \approx y

(very small),

[S^{2 -}] \approx 0.1 - 0.03 = 0.07 M

K_{c} = \frac{1}{[ Fe ^{2 +} ] [ S ^{2 -} ]} = 1.6 \times 1 0^{17}

[Fe^{2 +}] = \frac{1}{1.6 \times 1 0 ^{17} \times 0.07} = \frac{1}{1.12 \times 1 0 ^{16}} \approx 8.93 \times 1 0^{- 17} M

Y \approx 8.93

. The source key accepts 8.70 to 9.10.

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Numerical answer