JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
The equilibrium constant \(K_{c}\) at \(298\, \mathrm{~K}\) for the reaction \(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}\) is \(100 .\) Starting with an equimolar solution with concentrations of \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) all equal to \(1\, \mathrm{M}\), the equilibrium concentration of \(D\) is \(.......\times 10^{-2}\) M. (Nearest integer)
- A \(18\)
- B \(182\)
- C \(45\)
- D \(18200\)
Answer & Solution
Correct Answer
(B) \(182\)
Step-by-step Solution
Detailed explanation
\(\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}: \mathrm{K}_{\mathrm{eq}}=100\) \(1 \mathrm{M}\quad 1 \mathrm{M} \quad 1 \mathrm{M} \quad1 \mathrm{M}\) First check direction of reversible reaction. Since…
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