Q35 / 225 · JEE Main 2003 · Chemistry

For the reaction equilibrium $\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})$ the concentrations of $\mathrm{N}_2 \mathrm{O}_4$ and $\mathrm{NO}_2$ at equilibrium are $4.8 \times 10^{-2}$ and $1.2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$ respectively. The value of $\mathrm{K}_{\mathrm{c}}$ for the reaction is

  • A
    $3 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1}$
  • Correct answer
    B
    $3 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$
  • C
    $3 \times 10^3 \mathrm{~mol} \mathrm{~L}^{-1}$
  • D
    $3.3 \times 10^2 \mathrm{~mol} \mathrm{~L}^{-1}$

Solution

$\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2 \mathrm{O}_4\right]}=\frac{\left[1.2 \times 10^{-2}\right]^2}{\left[4.8 \times 10^{-2}\right]}=3 \times 10^{-3} \mathrm{~mol} / \mathrm{L}$

Hence, the correct answer is option: 'B'.

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