$K_{c}$ for the reaction $\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) + \frac{1}{2} B r_{2} (g) ⇌ N O B r (g)$ the following information at $298 K$
$2 N O (g) ⇌ N_{2} (g) + O_{2} (g) K_{1} = 2.4 \times 10^{30}$
$N O (g) + \frac{1}{2} B r_{2} (g) ⇌ N O B r (g) K_{2} = 1.4$
(Given : $\sqrt{\frac{1}{2.4}} = 0.645$ )

3.15 \times 10^{- 9}

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6.35 \times 10^{- 18}

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9.0 \times 10^{- 16}

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17 \times 10^{- 17}

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Solution

The correct option is C $9.0 \times 10^{- 16}$
$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ N O (g)$
$K_{c} = {(\frac{1}{2.4 \times 10^{30}})}^{\frac{1}{2}}$
$N O (g) + \frac{1}{2} B r_{2} (g) ⇌ N O B r (g)$
$K_{2} = 1.4$
$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) + \frac{1}{2} B r (g) ⇌ N O B r (g)$
$K = K_{c} \times K_{2} = 0.645 \times 1.4 \times 10^{- 15} = 0.903 \times 10^{- 15} \approx 9.0 \times 10^{- 16}$

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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 :