alyssa911DFTBA
Answered

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?

A2.1 x 10-2
B2.1 x 102
C1.2 x 103
D1.2 x 10-3

Verified Answer :

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) \(2.1\times 10^{-2}\)

Solution : Given,

Concentration of \(HF\) = \(5.82\times 10^{-2}M\)

Concentration of \(H_2\) = \(8.4\times 10^{-3}M\)

Concentration of \(F_2\) = \(8.4\times 10^{-3}M\)

The given balanced equilibrium reaction is,

\(2HF(g)\rightleftharpoons H_2(g)+F_2(g)\)

The expression for equilibrium constant will be,

\(K_{eq}=\frac{[H_2]\times [F_2]}{[HF]^2}\)

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

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

\(K_{eq}=2.08\times 10^{-2}=2.1\times 10^{-2}\)

Therefore, the value of equilibrium constant is, \(2.1\times 10^{-2}\)

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