The thermal decomposition of $N_{2} O_{4}$ g) $⇋$ 2 $N O_{2}$ (g) is as shown below:

$\begin{matrix} N_{2} O_{4} (g) ⇌ 2 N O_{2} (g) A t s t a r t & a m o l 0 m o l A t e q u b . & (a - x) m o l 2 x m o l \end{matrix}$

Where x = degree of dissociation. If V and P are equilibrium volume and pressure respectively then the corresponding expressions for K_c and K_p are $\frac{4 x^{2}}{V (a - x)}$ and $\frac{4 x^{2}}{a^{2} - x^{2}}$ (respectively). True or False?

True

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False

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Solution

The correct option is A
True

$Calculating K_{c} : \begin{matrix} N_{2} O_{4} (g) ⇌ 2 N O_{2} (g) A t s t a r t & a m o l 0 m o l A t e q u b . & (a - x) m o l 2 x m o l \end{matrix}$
$K_{c} = \frac{{[\frac{2 x}{V}]}^{2}}{\frac{a - x}{V}} = \frac{4 x^{2}}{(a - x) V}$

$Calculating K_{p} : \begin{matrix} N_{2} O_{4} (g) ⇌ 2 N O_{2} (g) A t s t a r t & a m o l 0 m o l A t e q u b . & (a - x) m o l 2 x m o l \end{matrix}$
$Total number of moles at equilibrium = a - x + 2 x = a + x$
$At equilibrium, P_{N O_{2}} = \frac{2 x + P}{a + x} P_{N_{2}} O_{4} = \frac{(a - x) \times P}{(a + x)} ∴ K_{p} = \frac{{(\frac{2 x \times P}{a + x})}^{2}}{\frac{(a - x) \times P}{a + x}} = \frac{4 x^{2} \times P}{(a - x) (a + x)} = \frac{4 x^{2} P}{a^{2} - x^{2}}$

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Similar questions

The thermal decomposition of $N_{2} O_{4}$ g) $⇋$ 2 $N O_{2}$ (g) is as shown below:

$\begin{matrix} N_{2} O_{4} (g) ⇌ 2 N O_{2} (g) A t s t a r t & a m o l 0 m o l A t e q u b . & (a - x) m o l 2 x m o l \end{matrix}$

Where x = degree of dissociation. If V and P are equilibrium volume and pressure respectively then the corresponding expressions for K_c and K_p are $\frac{4 x^{2}}{V (a - x)}$ and $\frac{4 x^{2}}{a^{2} - x^{2}}$ (respectively). True or False?

The thermal decomposition of $N_{2} O_{4}$ g) $⇋$ 2 $N O_{2}$ (g) is as shown below:

$\begin{matrix} N_{2} O_{4} (g) ⇌ 2 N O_{2} (g) A t s t a r t & a m o l 0 m o l A t e q u b . & (a - x) m o l 2 x m o l \end{matrix}$

For $N_{2} O_{4}$ (g) $⇋$ $2 N O_{2}$ (g),the value of degree of dissociation as a function of equilibrium pressure P and equilibrium constant $K_{p}$ is $α = \sqrt{\frac{K_{p}}{(4 p + K_{p})}}$ True or false?