For the system 3A + 2B ⇌ C, the expression for equilibrium constant is
Chemistry · JEE Main · NTA Exams — Equilibrium
For the system 3A + 2B \(\text { ⇌ }\) C, the expression for equilibrium constant is
- \(\frac{[3 \mathrm{~A}][2 \mathrm{~B}]}{\mathrm{C}}\)
- \(\frac{[\mathrm{C}]}{[3 \mathrm{~A}][2 \mathrm{~B}]}\)
- \(\frac{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}{[\mathrm{C}]}\)
- \(\frac{[\mathrm{C}]}{[\mathrm{A}]^{3}[\mathrm{~B}]^{2}}\)
Answer
(D) [ C ] [ A ]^ 3 [ ~B ]^ 2
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