Q35 / 160 · AP EAMCET 2009 · Chemistry

The equilibrium constant for the given reaction is 100 . $\mathrm{N}_2(g)+2 \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$ What is the equilibrium constant for the reaction given below? $\mathrm{NO}_2(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(g)+\mathrm{O}_2(g)$

  • A
    10
  • B
    1
  • Correct answer
    C
    0.1
  • D
    0.01

Solution

$\begin{gathered}\mathrm{N}_2+2 \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}_2 \\ K_1=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]^2}\end{gathered}$ or $\quad 100=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]^2} \quad \ldots$ (i) Again, $\begin{aligned} {\left[\mathrm{NO}_2\right] } & \rightleftharpoons \frac{1}{2} \mathrm{~N}_2+\mathrm{O}_2 \\ K_2 & =\frac{\left[\mathrm{N}_2\right]^{1 / 2}\left[\mathrm{O}_2\right]}{\left[\mathrm{NO}_2\right]}\end{aligned}$ Eqs. (i) $\times$ (ii), we get $\begin{gathered} 100 \times K_2^2=1 \\ \text { or } \quad K_2^2=\frac{1}{100} \quad \text { or } \quad K_2=\frac{1}{10}=0.1 \end{gathered}$

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

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