0.30 mole of N 2 O 4 is allowed to equilibrate in a 4 litre vessel. If x mole of N 2 O 4…
Chemistry · JEE Main · NTA Exams — Equilibrium
0.30 mole of N2O4 is allowed to equilibrate in a 4 litre vessel. If x mole of N2O4 dissociates and 0.4 mole of NO2 is present at equilibrium in the reaction, \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \div 2 \mathrm{NO}_{2}(\mathrm{~g})\), which of the following is the correct expression for the equilibrium constant?
- \(\mathbf{K}=\frac{\mathbf{x}^{2}}{1.2-4 \mathbf{x}}\)
- \(K=\frac{2 x^{2}}{0.3-x}\)
- \(\mathrm{K}=\frac{\mathbf{x}^{2}}{1.2-2 \mathbf{x}}\)
- \(K=\frac{x^{2}}{0.3-x}\)
Answer
(C) K = x ^ 2 1.2-2 x
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