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

Answer:

A). 2.1 × $10^{-2}$

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.