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Morgarella [4.7K]
3 years ago
7

At 500K, the equilibrium constant for the reaction N2O4D2NO2 is 1.5 x 10^3. What is the equilibrium constant for the reaction: 6

NO2 D 3N2O4?
Chemistry
1 answer:
Veronika [31]3 years ago
3 0

Answer:

K = 2.96x10⁻¹⁰

Explanation:

Based on the initial reaction:

N2O4 ⇄ 2NO2; K = 1.5x10³

Using Hess's law, we can multiply this reaction changing K:

3 times this reaction:

3N2O4 ⇄ 6NO2; K = (1.5x10³)³ =3.375x10⁹

The inverse reaction has a K of:

6NO2 ⇄ 3N2O4 K = 1/3.375x10⁹;

<h3>K = 2.96x10⁻¹⁰</h3>

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Sonja [21]

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M  respectively.

Solution :  Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

    H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially               0.11   0.11            C

At equilibrium  (0.11-x) (0.11-x)   (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M  = (0.11-x)

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

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