Kc for is x-10-16litre1-2mol-1-2- 1-2N2g - 1-2O2g - 1-2Br2g - NOBrgGiven that- 2NOg -N2g - O2g- Kc1 - 2-4-1030 NOg - 1-2Br2g - NOBrg- Kc2 - 1-4 litre1-2mol-1-2- 100x is-

2NO(g) ⇌N_2(g) + O_2(g) nbsp; ......(1). nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; ...

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Byju's Answer

Standard XII

Mathematics

One - One function

Kc for is x...

Question

$K_{c}$ for is $x \times 10^{- 16} {l i t r e}^{1 / 2} {m o l}^{- 1 / 2}$ ,
$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) + \frac{1}{2} B r_{2} (g) ⇌ N O B r (g)$
Given that,
$2 N O (g) ⇌ N_{2} (g) + O_{2} (g)$ ; $K_{c_{1}} = 2.4 \times 10^{30}$
$N O (g) + \frac{1}{2} B r_{2} (g) ⇌ N O B r (g)$ ; $K_{c_{2}} = 1.4 {l i t r e}^{1 / 2} {m o l}^{- 1 / 2}$ ]
$100 x$ is__________.

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Solution

$2 N O (g) ⇌ N_{2} (g) + O_{2} (g)$ ......(1).
$K_{c_{1}} = 2.4 \times 10^{30}$
$N O (g) + \frac{1}{2} B r_{2} (g) ⇌ N O B r (g)$ ......(2)
$K_{c_{2}} = 1.4 {l i t r e}^{1 / 2} {m o l}^{- 1 / 2}$ ]
Divide first reaction with 2.
$N O (g) ⇌ \frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g)$ ......(3).
Subtract third reaction from second.
$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) + \frac{1}{2} B r_{2} (g) ⇌ N O B r (g)$ ......(4)
$K_{c 3} = \sqrt{K_{c 1}} = \sqrt{2.4 \times 10^{30}} = 1.549 \times 10^{15}$ .
$K_{c 4} = \frac{K_{c 2}}{K_{c 3}} = \frac{1.4 {l i t r e}^{1 / 2} {m o l}^{- 1 / 2}}{1.549 \times 10^{15}} = 9.03 \times 10^{- 16} {l i t r e}^{1 / 2} {m o l}^{- 1 / 2} = x \times 10^{- 16} {l i t r e}^{1 / 2} {m o l}^{- 1 / 2}$
$x = 9.03$
$100 x = 903$ .

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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 :