JEE Main 2022 — Chemistry Physical Chemistry

2022integereasy

At $298K$ , the equilibrium constant is $2\times {10}^{15}$ for the reaction:

$\mathrm{Cu}(s)+2{\mathrm{Ag}}^{+}(\mathrm{aq})\rightleftharpoons {\mathrm{Cu}}^{2+}(\mathrm{aq})+2\mathrm{Ag}(s)$

The equilibrium constant for the reaction

$\frac{1}{2}{\mathrm{Cu}}^{2+}(\mathrm{aq})+\mathrm{Ag}(s)\rightleftharpoons \frac{1}{2}\mathrm{Cu}(s)+{\mathrm{Ag}}^{+}(\mathrm{aq})$ is $x\times {10}^{-8}$ . The value of $x$ is ______. (Round off the answer to the nearest integer)

Official previous-year question

Held on 26 Jul 2022 · Verified 6 Jul 2026.

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Solution

$\mathrm{Cu}(s)+2{\mathrm{Ag}}^{+}(\mathrm{aq})\rightleftharpoons {\mathrm{Cu}}^{2+}(\mathrm{aq})+2\mathrm{Ag}(s);{K}_{\mathrm{eq}}=2\times {10}^{15}$

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

The equilibrium reaction is reversed and multiplied with $\frac{1}{2}$ ,

Hence, ${K}_{\mathrm{eq}}^{'}=\frac{1}{\sqrt{{K}_{\mathrm{eq}}}}=\frac{1}{\sqrt{2\times {10}^{15}}}=x\times {10}^{-8}$

$\Rightarrow \frac{1}{\sqrt{20}}\times \frac{1}{{10}^{7}}=x\times {10}^{-8}$

$\Rightarrow \frac{1}{\sqrt{20}}\times {10}^{-7}=x\times {10}^{-8}$

$\frac{10}{\sqrt{20}}=x$

$\Rightarrow x=\frac{\sqrt{10}}{\sqrt{2}}=\sqrt{5}=2.236$

$\simeq 2.24$

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JEE Main 2022 — Chemistry Physical Chemistry

2022integereasy

At $298K$ , the equilibrium constant is $2\times {10}^{15}$ for the reaction:

$\mathrm{Cu}(s)+2{\mathrm{Ag}}^{+}(\mathrm{aq})\rightleftharpoons {\mathrm{Cu}}^{2+}(\mathrm{aq})+2\mathrm{Ag}(s)$

The equilibrium constant for the reaction

Official previous-year question

Held on 26 Jul 2022 · Verified 6 Jul 2026.

Did you get this right?

Solution

$\mathrm{Cu}(s)+2{\mathrm{Ag}}^{+}(\mathrm{aq})\rightleftharpoons {\mathrm{Cu}}^{2+}(\mathrm{aq})+2\mathrm{Ag}(s);{K}_{\mathrm{eq}}=2\times {10}^{15}$

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

The equilibrium reaction is reversed and multiplied with $\frac{1}{2}$ ,

Hence, ${K}_{\mathrm{eq}}^{'}=\frac{1}{\sqrt{{K}_{\mathrm{eq}}}}=\frac{1}{\sqrt{2\times {10}^{15}}}=x\times {10}^{-8}$

$\Rightarrow \frac{1}{\sqrt{20}}\times \frac{1}{{10}^{7}}=x\times {10}^{-8}$

$\Rightarrow \frac{1}{\sqrt{20}}\times {10}^{-7}=x\times {10}^{-8}$

$\frac{10}{\sqrt{20}}=x$

$\Rightarrow x=\frac{\sqrt{10}}{\sqrt{2}}=\sqrt{5}=2.236$

$\simeq 2.24$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

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JEE Main 2022 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2022 — Chemistry Physical Chemistry

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry