Determine $K_{c}$ for the reaction, $\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) + \frac{1}{2} B r_{2} (g) ⇌ N O B r (g)$ from the following data at $298 K$ ;
The equilibrium constants for the following reactions,
$2 N O (g) ⇌ N_{2} (g) + O_{2} (g)$ and $N O (g) + \frac{1}{2} B r_{2} (g) ⇌ N O B r (g)$
$are 0.36 \times 10^{30} and 1.8 respectively :$

3.0 \times 10^{- 15}

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4.5 \times 10^{- 14}

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9.1 \times 10^{- 15}

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1.9 \times 10^{- 14}

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Solution

The correct option is A $3.0 \times 10^{- 15}$
The net reaction is,
$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) + \frac{1}{2} B r_{2} (g) ⇌ N O B r (g)$
$K_{c_{n e t}} = \frac{[N O B r]}{[N_{2}]^{1 / 2} [O_{2}]^{1 / 2} [B r_{2}]^{1 / 2}}$
Considering the given equations:
$2 N O (g) ⇌ N_{2} (g) + O_{2} (g); K_{c} = 0.36 \times 10^{30}$
$∴ N_{2} (g) + O_{2} (g) ⇌ 2 N O (g); K_{c}^{/} = \frac{1}{0.36 \times 10^{30}}$
$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ N O (g); K_{c}^{/ /} = {(\frac{1}{0.36 \times 10^{30}})}^{1 / 2}$

$∴ \frac{[N O]}{[N_{2}]^{1 / 2} [O_{2}]^{1 / 2}} = K_{c}^{/ /} = {(\frac{1}{0.36 \times 10^{30}})}^{1 / 2} \cdot \cdot \cdot (i)$

Again,
$N O (g) + \frac{1}{2} B r_{2} (g) ⇌ N O B r (g) K_{c}^{/ / /} = 1.8$

So, $\frac{[N O B r]}{[N O] [B r_{2}]^{1 / 2}} = K_{c}^{/ / /} = 1.8 \cdot \cdot \cdot (i i)$
Multiplying both equations
$\frac{[N O]}{[N_{2}]^{1 / 2} [O_{2}]^{1 / 2}} \times \frac{[N O B r]}{[N O] [B r_{2}]^{1 / 2}} = K_{c}^{/ /} \times K_{c}^{/ / /} = {(\frac{1}{0.36 \times 10^{30}})}^{1 / 2} \times 1.8$
$\Rightarrow \frac{[N O B r]}{[N_{2}]^{1 / 2} [B r_{2}]^{1 / 2}} = K_{c_{n e t}} = 3 \times 10^{- 15}$

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Similar questions

Q. Determine

K_{c}

for the reaction

\frac{1}{2} N_{2} (g) + \frac{3}{2} O_{2} (g) + \frac{1}{2} {B r}_{2} ((g)) ⇌ N O B r (g)

from the following data at

298 K

The equilibrium constants for the following reactions,

2 N O (g) ⇌ N_{2} (g) + O_{2} (g)

and

2 N O (g) ⇌ \frac{1}{2} {B r}_{2} ((g)) ⇌ N O B r (g)

are

2.4 \times 10^{30}

and

1.4

respectively.

Consider the following equilibria at $25^{\circ} C, 2 N O_{(g)} ⇌ N_{2 (g)} + O_{2 (g)}; K_{1} = 4 \times 10^{30} a n d N O_{(g)} + \frac{1}{2} B r_{2 (g)} ⇌ N O B r_{(g)}; K_{2} = 1.4 m o l^{- \frac{1}{2}} L^{\frac{1}{2}} .$ The value of K_C for the reaction (at same temperature) $\frac{1}{2} N_{2 (g)} + \frac{1}{2} O_{2 (g)} + \frac{1}{2} B r_{2 (g)} ⇌ N O B r_{(g)}$ is :