Question

Consider the following reactions and their respective equilibrium constants: NO(g)+12Br2(g)⇌NOBr(g)Kp=5.3 2NO(g)⇌N2(g)+O2(g)Kp=2.1×1030 Use these reactions and their equilibrium constants to predict the equilibrium constant for the following reaction: N2(g)+O2(g)+Br2(g)⇌2NOBr(g)

Answer 1

Answer:

Equilibrium constant of the given reaction is $1.3\times 10^{-29}$

Explanation:

$NO+\frac{1}{2}Br_{2}\rightleftharpoons NOBr$....$(K_{p})_{1}=5.3$

$2NO\rightleftharpoons N_{2}+O_{2}$....$(K_{p})_{2}=2.1\times 10^{30}$

The given reaction can be written as summation of the following reaction-

$2NO+Br_{2}\rightleftharpoons 2NOBr$

$N_{2}+O_{2}\rightleftharpoons 2NO$

......................................................................................

$N_{2}+O_{2}+Br_{2}\rightleftharpoons 2NOBr$

Equilibrium constant of this reaction is given as-

$\frac{[NOBr]^{2}}{[N_{2}][O_{2}][Br_{2}]}$

$=(\frac{[NOBr]}{[NO][Br_{2}]^{\frac{1}{2}}})^{2}(\frac{[NO]^{2}}{[N_{2}][O_{2}]})$

$=\frac{(K_{p})_{1}^{2}}{(K_{p})_{2}}$

$=\frac{(5.3)^{2}}{2.1\times 10^{30}}=1.3\times 10^{-29}$

Answer 2

Answer :  The equilibrium constant for the given reaction is, $1.3\times 10^{-29}$

Explanation :

The given equilibrium reactions are,

(i) $NO(g)+\frac{1}{2}Br_2(g)\rightleftharpoons NOBr(g)$;   $K_p_1=5.3$

(ii) $2NO(g)\rightleftharpoons N_2(g)+O_2(g)$;   $K_p_2=2.1\times 10^{30}$

Now we have to determine the equilibrium constants for the following equilibrium reactions.

(iii) $N_2(g)+O_2(g)+Br_2(g)\rightleftharpoons 2NOBr(g)$;   $K_p=?$

From the given reaction we conclude that, the reaction (iii) will takes place when reverse the reaction (ii) and reaction (i) is multiplied by 2 and then adding all the reaction.

Thus, the equilibrium constant will be:

$K_p=\frac{1}{K_p_2}\times (K_p_1)^2$

Now put all the given values in this expression, we get:

$K_p=\frac{1}{2.1\times 10^{30}}\times (5.3)^2$

$K_p=1.3\times 10^{-29}$

Therefore, the equilibrium constant for the given reaction is, $1.3\times 10^{-29}$