MCQ
For the reaction in equilibrium
$2NOB{r_{(g)\,}}\, \rightleftharpoons \,2N{O_{(g)}}\, + \,B{r_{2(g)}}$
if ${P_{B{r_2}}}$ is $\frac {P}{4}$ At equilibrium and $P$ is total pressure then calculate $\frac {P}{K_P}.$
- ✓$1$
- B$4$
- C$4/9$
- D$2.25$
✓
Answer
Correct option: A.
$1$
a
$\mathop {2NOB{r_{(g)}}}\limits_{P - \frac{{3P}}{4}} \rightleftharpoons \mathop {2N{O_{(g)}}}\limits_{\frac{{2P}}{4}} + \mathop {B{r_{2(g)}}}\limits_{\frac{P}{4}} $
$\mathop {2NOB{r_{(g)}}}\limits_{P - \frac{{3P}}{4}} \rightleftharpoons \mathop {2N{O_{(g)}}}\limits_{\frac{{2P}}{4}} + \mathop {B{r_{2(g)}}}\limits_{\frac{P}{4}} $
$P_{T}=P-\frac{3 P}{4}+\frac{2 P}{4}+\frac{P}{4}=P$
$K_{P}=\frac{\left(\frac{2 P}{4}\right)^{2}\left(\frac{P}{4}\right)}{\left(\frac{P}{4}\right)^{2}}=\frac{4}{9} \times \frac{P}{4}=\frac{P}{1}$
$\boxed{\frac{P}{{{K_P}}} = 1}$
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