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