Answer :
Answer: The molar concentration of HF at equilibrium is [tex]5.64\times 10^{-3}[/tex]
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] ......(1)
- For hydrogen gas:
Molarity of hydrogen gas solution = 0.0052 M
Volume of solution = 3.9 L
Putting values in equation 1, we get:
[tex]0.0052M=\frac{\text{Moles of }H_2}{3.9L}\\\\\text{Moles of }H_2=(0.0052mol/L\times 3.9L)=0.0203mol[/tex]
- For fluorine:
Molarity of fluorine solution = 0.034 M
Volume of solution = 3.3 L
Putting values in equation 1, we get:
[tex]0.034M=\frac{\text{Moles of }F_2}{3.3L}\\\\\text{Moles of }F_2=(0.034mol/L\times 3.3L)=0.1122mol[/tex]
For the given chemical equation:
[tex]H_2(g)+F_2(g)\rightleftharpoons 2HF(g)[/tex]
Initial: 0.0203 0.1122
At eqllm: 0.0203-x 0.1122-x 2x
Volume of the solution = [3.3 + 3.9] L = 7.2 L
Relation of [tex]K_p[/tex] with [tex]K_c[/tex] is given by the formula:
[tex]K_p=K_c(RT)^{\Delta ng}[/tex]
where,
[tex]K_p[/tex] = equilibrium constant in terms of partial pressure = [tex]7.8\times 10^{14}[/tex]
[tex]K_c[/tex] = equilibrium constant in terms of concentration = ?
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature
[tex]\Delta n_g[/tex] = change in number of moles of gas particles = [tex]n_{products}-n_{reactants}=2-2=0[/tex]
Putting values in above equation, we get:
[tex]7.8\times 10^{14}=K_c\times (0.0821\times T)^{0}\\\\K_c=7.8\times 10^{14}[/tex]
The expression of K_c for above equation follows:
[tex]K_c=\frac{[HF]^2}{[F_2][H_2]}[/tex]
We are given:
[tex]K_c=7.8\times 10^{14}[/tex]
[tex][HF]_{eq}=\frac{2x}{7.2}[/tex]
[tex][H_2]_{eq}=\frac{(0.0203)}{7.2}[/tex]
[tex][F_2]_{eq}=\frac{(0.1122-x)}{7.2}[/tex]
Putting values in above expression, we get:
[tex]7.8\times 10^{14}=\frac{(2x/7.2)^2}{((0.0203-x)/7.2)\times ((0.1122-x)/7.2)}\\\\x=0.0203,0.1122[/tex]
Neglecting the value of x = 0.1122 because equilibrium concentration of hydrogen gas will become negative, which is not possible
So, equilibrium moles of [tex]HF=2x=(2\times 0.0203)=0.0406moles[/tex]
Now, calculating the molarity of HF by using equation 1,
Moles of HF = 0.0406 moles
Volume of solution = 7.2 L
Putting values in equation 1, we get:
[tex]\text{Molarity of HF}=\frac{0.0406mol}{7.2L}=5.64\times 10^{-3}[/tex]
Hence, the molar concentration of HF at equilibrium is [tex]5.64\times 10^{-3}[/tex]