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QJEEM22Ch436e: JEE Main 2022 Chemistry - Class 11 JEEM22Ch436e Solution: At , the equilibrium constant is for the reaction : The…

At 298 K298 \mathrm{~K} , the equilibrium constant is 2×10152 \times 10^{15} 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})

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})

is x×10−8x \times 10^{-8} . The value of xx is _____________. (Nearest Integer)

Step-by-Step Verified Solution

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})

k=2×1015k=2 \times 10^{15}

12Cu(s)+Ag+(aq)⇌Cu+2(aq)+2Ag(s)\frac{1}{2} \mathrm{Cu}(\mathrm{s})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightleftharpoons \mathrm{Cu}^{+2}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})

K′=1( K)1/2=1(2×1015)1/2\mathrm{K}^{\prime}=\frac{1}{(\mathrm{~K})^{1 / 2}}=\frac{1}{\left(2 \times 10^{15}\right)^{1 / 2}}

=2.23×10−8=2.23 \times 10^{-8}

x≃2x \simeq 2

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