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.

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Solution

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;
$2 N O (g) ⇌ N_{2} (g) + O_{2} (g);$ equilibrium constant $= 2.4 \times 10^{30}$
Exchange the reactants and products of above equation and take reciprocal of the equilibrium constant
$N_{2} (g) + O_{2} (g) ⇌ 2 N O (g);$ equilbrium constant $= (\frac{1}{2.4 \times 10^{30}})$
Divide above equation with 2 and take square root of equilibrium constant expression.
$\frac{1}{2} N_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ 2 N O (g)$
Equilibrium constant $= {(\frac{1}{2.4 \times 10^{30}})}^{1 / 2} = 0.6455 \times 10^{- 15}$
$\frac{[N O]}{{[N_{2}]}^{1 / 2} {[O_{2}]}^{1 / 2}} = K_{c}^{'} = 0.6455 \times 10^{- 15} . . . (i)$
$N O (g) + \frac{1}{2} {B r}_{2} (g) ⇌ N O B r (g)$
$\frac{[N O B r]}{[N O] {[{B r}_{2}]}^{1 / 2}} = K_{c}^{''} = 1.4... (i i)$
Multiplying equations ( i) and (ii)
$\frac{[N O]}{{[N_{2}]}^{1 / 2} {[O_{2}]}^{1 / 2}} \times \frac{[N O B r]}{[N O] {[{B r}_{2}]}^{1 / 2}} = \frac{[N O B r]}{{[N_{2}]}^{1 / 2} {[O_{2}]}^{1 / 2} {[{B r}_{2}]}^{1 / 2}}$
$K_{c (n e t)} = K_{c}^{'} \times K_{c}^{''} = 0.6455 \times 10^{- 15} \times 1.4 = 9.037 \times 10^{- 16}$

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Similar questions

Q. 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 :

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

)

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 :

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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Determine Kc for the reaction12N2(g)+32O2(g)+12Br2((g))⇌NOBr(g) from the following data at 298KThe equilibrium constants for the following reactions,2NO(g)⇌N2(g)+O2(g) and 2NO(g)⇌12Br2((g))⇌NOBr(g) are 2.4×1030 and 1.4 respectively.