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Chemical Equilibrium question

2022 · 26 Jul · Shift 1 · Q20
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  5. /2022 · 26 Jul · Shift 1 · Q20

Chemical Equilibrium question

2022 · 26 Jul · Shift 1 · Q20

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
At 298 K298 \mathrm{~K}298 K, the equilibrium constant is 2×10152 \times 10^{15}2×1015 for the reaction : Cu(s)+2Ag+(aq)⇌Cu2+(aq)+2Ag(s)\mathrm{Cu}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})Cu(s)+2Ag+(aq)⇌Cu2+(aq)+2Ag(s) The equilibrium constant for the reaction 12Cu2+(aq)+Ag(s)⇌12Cu(s)+Ag+(aq)\frac{1}{2} \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s}) \rightleftharpoons \frac{1}{2} \mathrm{Cu}(\mathrm{s})+\mathrm{Ag}^{+}(\mathrm{aq})21​Cu2+(aq)+Ag(s)⇌21​Cu(s)+Ag+(aq) is x×10−8x \times 10^{-8}x×10−8. The value of xxx is ‾\underline{\hspace{2cm}}​. (Nearest Integer)
Numerical answer
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Correct answer: 2

  1. Let the given reaction be Cu(s)+2Ag+(aq)⇌Cu2+(aq)+2Ag(s)\mathrm{Cu}(s)+2\mathrm{Ag}^+(aq) \rightleftharpoons \mathrm{Cu}^{2+}(aq)+2\mathrm{Ag}(s)Cu(s)+2Ag+(aq)⇌Cu2+(aq)+2Ag(s) with equilibrium constant K1=2×1015K_1=2\times 10^{15}K1​=2×1015

  2. We need the equilibrium constant for: 12Cu2+(aq)+Ag(s)⇌12Cu(s)+Ag+(aq)\frac{1}{2}\mathrm{Cu}^{2+}(aq)+\mathrm{Ag}(s) \rightleftharpoons \frac{1}{2}\mathrm{Cu}(s)+\mathrm{Ag}^+(aq)21​Cu2+(aq)+Ag(s)⇌21​Cu(s)+Ag+(aq) Call this constant K2K_2K2​.

  3. Observe how the target reaction is related to the given reaction.

  • First, reverse the given reaction: Cu2+(aq)+2Ag(s)⇌Cu(s)+2Ag+(aq)\mathrm{Cu}^{2+}(aq)+2\mathrm{Ag}(s) \rightleftharpoons \mathrm{Cu}(s)+2\mathrm{Ag}^+(aq)Cu2+(aq)+2Ag(s)⇌Cu(s)+2Ag+(aq) For reversal, K=1K1=12×1015K=\frac{1}{K_1}=\frac{1}{2\times 10^{15}}K=K1​1​=2×10151​

  • Now divide the entire reversed reaction by 222: 12Cu2+(aq)+Ag(s)⇌12Cu(s)+Ag+(aq)\frac{1}{2}\mathrm{Cu}^{2+}(aq)+\mathrm{Ag}(s) \rightleftharpoons \frac{1}{2}\mathrm{Cu}(s)+\mathrm{Ag}^+(aq)21​Cu2+(aq)+Ag(s)⇌21​Cu(s)+Ag+(aq) This is exactly the required reaction.

When coefficients are multiplied by 12\frac1221​, the equilibrium constant becomes the square root: K2=(12×1015)1/2K_2=\left(\frac{1}{2\times 10^{15}}\right)^{1/2}K2​=(2×10151​)1/2

  1. Calculate K2K_2K2​: K2=12×1015K_2=\frac{1}{\sqrt{2\times 10^{15}}}K2​=2×1015​1​ =12×107.5=\frac{1}{\sqrt{2}\times 10^{7.5}}=2​×107.51​ Using 107.5=10×10710^{7.5}=\sqrt{10}\times 10^7107.5=10​×107, K2=120×107K_2=\frac{1}{\sqrt{20}\times 10^7}K2​=20​×1071​ Now, 20≈4.472\sqrt{20}\approx 4.47220​≈4.472 So, K2≈14.472×107K_2\approx \frac{1}{4.472\times 10^7}K2​≈4.472×1071​ K2≈2.236×10−8K_2\approx 2.236\times 10^{-8}K2​≈2.236×10−8

  2. Given that K2=x×10−8K_2=x\times 10^{-8}K2​=x×10−8 we get x≈2.236x\approx 2.236x≈2.236 Nearest integer: x=2x=2x=2

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