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 :  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, $1.3\times 10^{-29}$

Answer 2

Considering the following reactions and their respective equilibrium constants: Using these reactions and their equilibrium constants to predict the equilibrium constant for the following reaction: The equilibrium constant for the following reaction is 1.3 × 10⁻²⁹

Equilibrium constant $\mathbf{K_p}$ indicates the proportion between a product versus the reactant in a reaction at equilibrium with regard to a particular unit.

From the given reaction;

$\mathbf{NO_{(g)} + \dfrac{1}{2}Br_{2(g)} \to NOBr_{(g)} \ \ \ \ \ \ \ \ K_p = 5.3 --- (1)}$

$\mathbf{2 NO_{(g)} = N_{2(g)} + O_{2(g)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ K_{p2} = 2.1 \times 10^{30} --- (2) }$

The equation for the above equilibrium constant can be expressed as follows:

$\mathbf{K_{P1} = \dfrac{P_{NOBr}}{P_{NO}\ P_{Br_2}^{1/2}}}$

$\mathbf{K_{P2} = \dfrac{P_{N_2} P_{O_2}}{P_{NO} \ ^2}}$

However, a required equation whose equilibrium was to be determined was given as:

$\mathbf{N_{2(g)}+O_{2(g)}+ Br_2_{(g)} \to 2 NOBr_{(g)}}$

The equilibrium constant is:

$\mathbf{K_p = \dfrac{P_{NOBr} \ ^2}{P_{N_2} P_{O_2} P_{Br_2} }}$

If we multiply equations (1) and (2) in a fashioned way such that equation one is squared and the reciprocal of (2) is derived, then we have:

$\mathbf{{(Kp_1)}^2 \times \dfrac{1}{K_P^2} =\Big ( \dfrac{P_{NOBr}}{P_{NO} \ P_{Br}^{1/2}}\Big)^2 \times \dfrac{P_{NO} \ ^2}{P_{N_2} \ P_{O_2}} }$

Then; we can say that:

$\mathbf{K_p \implies \dfrac{P_{NOBr} \ ^2}{P_{N_2} \ P_{O_2} \ P_{Br}} }$

Hence, the equilibrium constant kp is similar to Kp₁, and Kp₂.

Then;

$\mathbf{K_p = (K_{P_1})^2 \times \dfrac{1}{Kp_2}}$

Replacing the given values for the Kp's,

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

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

Therefore, we can conclude that the equilibrium constant for the following reaction is 1.3 × 10⁻²⁹.

Learn more about the equilibrium constant here:

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