Question

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

Answer 1

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 2

Answer : The correct option is, (A)
`2.1* 10^(-2)`

Solution : Given,

Concentration of
`HF` =
`5.82* 10^(-2)M`

Concentration of
`H_2` =
`8.4* 10^(-3)M`

Concentration of
`F_2` =
`8.4* 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)=([H_2]* [F_2])/([HF]^2)`

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

`K_(eq)=((8.4* 10^(-3))* (8.4* 10^(-3)))/((5.82* 10^(-2))^2)`

`K_(eq)=2.08* 10^(-2)=2.1* 10^(-2)`

Therefore, the value of equilibrium constant is,
`2.1* 10^(-2)`