At 298 ~K, the equilibrium constant is 2 × 10¹⁵ for the reaction: Cu(s)+2 Ag⁺(aq) Cu²⁺(aq)+2 Ag(s) The equilibrium constant for the reaction 1/2 Cu²⁺(aq)+Ag(s) 1/2 Cu(s)+Ag⁺(aq) is x × 10⁻⁸. The value of x is _ _ _ _ _ _ _. (Nearest Integer)

Equilibrium: JEE Main 2022 Chemistry question

JEE Main 2022 · 26 Jul, Morning shift · Q4

JEE Main 202226 Jul, Morning shiftChemistryHardEquilibrium123× since 2002

298 \mathrm{~K}

, the equilibrium constant is

2 \times 10^{15}

for the reaction :

\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

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

x \times 10^{-8}

. The value of

x

is _____________. (Nearest Integer)

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2022 Equilibrium Question with Answer (26 Jul Morning shift, Q4) | JEE Arc