Question

For the reaction below at a certain temperature, it is found that the equilibrium concentrations in a 5.00 L rigid container are [H2] = 0.0500 M, [F2] = 0.0100 M, and [HF] = 0.400 M. If 0.345 mol of F2 is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished in moles/liter.H2(g) + F2(g) <--->2 HF(g)

Answer 1

Answer:

$[HF]_{eq}=0.469M, [H_2]_{eq}=0.0155M, [F_2]_{eq}=0.0445M$

Explanation:

Hello,

At the first condition, one computes the equilibrium constant based on the law of mass action:

$K_{eq}=\frac{[HF]^2}{[H_2][F_2]}=\frac{(0.400M)^2}{(0.0500M)(0.0100M)} =320$

Now, we compute the final molarity of fluorine after the mentioned addition of 0.345 mol:

$M_{F_2}^{new}=\frac{5.00L*0.0100\frac{molF_2}{L}+0.354mol}{5.00L} =0.0790M$

Now, one proposes the law of mass action in therms of the affection of the equilibrium concentrations considering there is no change in the equilibrium constant:

$K_{eq}=\frac{(0.400M-2x)^2}{(0.0500M-x)(0.0790M-x)}$

$K_{eq}=\frac{0.16+1.6x+4x^2}{0.00395-0.129x+x^2}\\(0.00395-0.129x+x^2)K_{eq}=0.16+1.6x+4x^2\\1.264-41.28x+320x^2=0.16+1.6x+4x^2\\316x^2-42.88x+1.104=0\\x_1=0.101M\\x_2=0.0345M$

Finally, one selects 0.0345M since the other value gives negative concentrations, thus, the new concentrations are:

$[HF]_{eq}=0.469M, [H_2]_{eq}=0.0155M, [F_2]_{eq}=0.0445M$

Best regards.

Answer 2

The study of chemicals and bonds is called chemistry. There are two types of elements and these are metals and nonmetals.

The correct answer is 0.0445

What are moles?

• The mole is the base unit of the amount of substance in the International System of Units.

• It is defined as exactly 6.02214076×10²³ elementary entities, which may be atoms, molecules, ions, or electrons

According to the question,

At the first condition, one computes the equilibrium constant based on the law of mass action:

$K_{eq}= \frac{[Hf]^2}{[H_2][F_2]}$

$=\frac{0.4^2}{0.05*0.01}$ =320

we compute the final molarity of fluorine after the mentioned addition of 0.345 mole:-

$K_{eq} =\frac{5*0.01+0.345}{5} \\=0.0790$

Now, one proposes the law of mass action in terms of the affection of the equilibrium concentrations considering there is no change in the equilibrium constant:

$K_{eq}=\frac{[0.04^2-2x]^2}{[0.050-x][0.0790-x]}$

After solving the equation, the value of x will be:-

$x_1= 0.101\\X_2=0.0345$

Finally, one selects 0.0345M since the other value gives negative concentrations, thus, the new concentrations are:

[HF]=0.469, [H2]= 0.0155, [F2]=0.0445M

For more information about the moles, refer to the link:-

https://brainly.com/question/16759172