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QJEEAdv19Ch960b: JEE Advanced 2019 Chemistry - Class 11 JEEAdv19Ch960b Solution: For the following reaction, the equilibrium constant K c at 298 …

For the following reaction, the equilibrium constant Kc at 298 K is 1.6 ×\times 1017.

Fe2+(aq) + S2-(aq) ⇌ FeS(s)

When equal volumes of

0.06 M Fe2+(aq) and 0.2 M S2 −- (aq)

solutions are mixed, the equilibrium concentration of Fe2+(aq) is found by Y ×\times 10 −- 17 M. The value of Y is .................

Step-by-Step Verified Solution

Given, equilibrium constant (Kc) at 298 K = 1.6 ×\times 1017

JEE Advanced 2019 Paper 1 Offline Chemistry - Chemical Equilibrium Question 5 English Explanation
[Here, Kc >>103, thus limiting reagent will be consumed almost completely, 0.03 −- X = 0 ∴\therefore X = 0.03]

From equilibrium constant,

Kc=[FeS][Fe2+][S2−]{K_c} = {{[FeS]} \over {[F{e^{2 + }}][{S^{2 - }}]}}

Kc=1X×0.07{K_c} = {1 \over {X \times 0.07}}

[For Fes(s)(Pure solid)=1 mol L−1]\mathop {Fes(s)}\limits_{(Pure\,solid)} = 1\,mol\,{L^{ - 1}}] ]

1.6×1017=1X×0.071.6 \times {10^{17}} = {1 \over {X \times 0.07}}

X=11.6×1017×0.07X = {1 \over {1.6 \times {{10}^{17}} \times 0.07}}

=8.9×10−17= 8.9 \times {10^{ - 17}}

Given, X = Y ×\times 10-17 = 8.9 ×\times 10-17

∴\therefore Y = 8.9

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