alyssa911DFTBA
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2HF(g)<-----> H2(g)+F2(g)

At equilibrium at 600 K, the concentrations are as follows.

[HF] = 5.82 x 10-2 M
[H2] = 8.4 x 10-3 M
[F2] = 8.4 x 10-3 M

What is the value of Keq for the reaction expressed in scientific notation?
A. 2.1 x 10-2
B. 2.1 x 102
C. 1.2 x 103
D. 1.2 x 10-3

Respuesta :

yashddsy26
just use the eqn ,

so Keq = (H2 ×F2)/HF^2

so , (8.4×10^-3 / 5.82 ×10^-2) ^2

therefore, answer is A !

if you have any confusion, or need any help , just comment !

Answer : The correct option is, (A) [tex]2.1\times 10^{-2}[/tex]

Solution : Given,

Concentration of [tex]HF[/tex] = [tex]5.82\times 10^{-2}M[/tex]

Concentration of [tex]H_2[/tex] = [tex]8.4\times 10^{-3}M[/tex]

Concentration of [tex]F_2[/tex] = [tex]8.4\times 10^{-3}M[/tex]

The given balanced equilibrium reaction is,

[tex]2HF(g)\rightleftharpoons H_2(g)+F_2(g)[/tex]

The expression for equilibrium constant will be,

[tex]K_{eq}=\frac{[H_2]\times [F_2]}{[HF]^2}[/tex]

Now put all the given values in this formula, we get

[tex]K_{eq}=\frac{(8.4\times 10^{-3})\times (8.4\times 10^{-3})}{(5.82\times 10^{-2})^2}[/tex]

[tex]K_{eq}=2.08\times 10^{-2}=2.1\times 10^{-2}[/tex]

Therefore, the value of equilibrium constant is, [tex]2.1\times 10^{-2}[/tex]