For a reversible reaction : A + B ⇌ C ( dx dt )=2.0 10^ 3 ~L ~mol ^ -1 ~s ^ -1 [ ~A ][ B…

Chemistry · JEE Advanced · NTA ExamsEquilibrium

For a reversible reaction : A + B \(\text { ⇌ }\) C
\(\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)=2.0 \times 10^{3} \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{~s}^{-1}[\mathrm{~A}][\mathrm{B}]-1.0 \times 10^{2} \mathrm{~s}^{-1}[\mathrm{C}]\)
where x is the amount of ‘A’ dissociated. The value of equilibrium constant (Keq) is :
  1. 10
  2. 0.05
  3. 20
  4. Can’t be calculated

Answer

(C) 20

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