MCQ
For the reaction $2NOBr (g)  \rightleftharpoons  2NO(g) + Br_2(g)$, if ${P_{B{r_2}}} = \frac{P}{9}$ at equilibrium and $P$ is total presssure, then find $\frac{{{K_P}}}{P}$ ?
  • A
    $\frac{1}{9}$
  • $\frac{1}{81}$
  • C
    $\frac{1}{27}$
  • D
    $\frac{1}{3}$

Answer

Correct option: B.
$\frac{1}{81}$
b
    $2NOBr(g) = 2NO(g) + B{r_2}(g)$

at equilibrium           $\frac{{2P}}{9}$               $\frac{{P}}{9}$

$\left\{ \begin{gathered}
  a + \frac{{2P}}{9} + \frac{P}{9} = P \hfill \\
  a = \frac{{2P}}{3} \hfill \\ 
\end{gathered}  \right.$

$K_{p}=\frac{\left(\frac{2 P}{9}\right)^{2}\left(\frac{P}{9}\right)}{\left(\frac{2 P}{3}\right)^{2}}=\frac{P}{81}$

$\boxed{\frac{{{K_p}}}{P} = \frac{1}{{81}}}$

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