For the following reaction, the equilibrium constant K c at 298 K is 1.6 × 10 17 Fe 2+ (aq) + S 2- (aq) rightleftharpoons 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 × 10 -17 M. The value of Y is ldots ldots ldots

Chemistry · Chemical Equilibrium

JEE Advanced 2019 — Paper 1 — Question 26

\text{For the following reaction, the equilibrium constant } K_c \text{ at } 298 \text{ K is } 1.6 \times 10^{17}

\text{Fe}^{2+}(aq) + \text{S}^{2-}(aq) \rightleftharpoons \text{FeS}(s)

\text{When equal volumes of } 0.06 \text{ M Fe}^{2+}(aq) \text{ and } 0.2 \text{ M S}^{2-}(aq) \text{ solutions are mixed, the }

\text{ equilibrium concentration of Fe}^{2+}(aq) \text{ is found to be } Y \times 10^{-17} \text{ M. The value of } Y \text{ is }

$\ldots \ldots \ldots$

Numerical answer — enter this value.

\text{Initial } [\text{Fe}^{2+}] = 0.06 \text{ M}, [\text{S}^{2-}] = 0.2 \text{ M}

\text{After mixing } [\text{Fe}^{2+}] = 0.03 \text{ M}, [\text{S}^{2-}] = 0.1 \text{ M}

\text{At equilibrium } [\text{Fe}^{2+}] = ?, [\text{S}^{2-}] \approx 0.07 \text{ M}

K_c = 1.6 \times 10^{17} = \frac{1}{[\text{Fe}^{2+}][\text{S}^{2-}]}

1.6 \times 10^{17} = \frac{1}{[\text{Fe}^{2+}] \times 0.07}

[\text{Fe}^{2+}] = \frac{1}{1.6 \times 10^{17} \times 0.07} = \frac{10^{-17}}{1.6 \times 0.07} = \frac{10^{-17}}{0.112}

[\text{Fe}^{2+}] = 8.928 \times 10^{-17} \text{ M}

\text{Given } [\text{Fe}^{2+}] = Y \times 10^{-17} \text{ M}

Y \times 10^{-17} = 8.928 \times 10^{-17}

Y \approx 8.93

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Exam: JEE Advanced 2019
Paper: Paper 1
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient