Question

Consider the reaction. 2HF(g)—H2(g)+F2(g). What is the value of Keq for the reaction expressed in scientific notation

Answer 1

A). 2.1 ×
`10^(-2)`

Step-by-step explanation:

Given reaction,

2
`HF` (g) ⇄
`H_(2)` (g) +
`F_(2)` (g)

The concentrations are as following;

`HF` = 5.82 ×
`10^(-2)` M

`H_(2)` = 8.4 ×
`10^(-3)` M

`F_(2)` = 8.4 ×
`10^(-3)` M

So,

`K_(eq)` = [(
`H_(2)` ) × (
`F_(2)`)] ÷ [
`HF`]^2

Now,

We can determine the value of
`K_(eq)` by substituting the values in above formula:

`K_(eq)` = [ (8.4 ×
`10^(-3)` M) × (8.4 ×
`10^(-3)` M)] ÷ [(5.82 ×
`10^(-2)`)^2

= 2.08 *
`10^(-2)`

= 2.1 ×
`10^(-2)`

∵
`K_(eq)` = 2.1 ×
`10^(-2)`

Thus, option A is the correct answer.