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.