Answer to Question #155409 in General Chemistry for C

Question

3. At 300 K, 0.045 mol of hydrogen and fluorine are placed in a closed 1.5 L container and allowed to reach equilibrium.

               H2(g) + F2(g) <---->  2HF(g)                     Keq = 4.3 at 300 K

What is the equilibrium concentration of HF(g)?  

Answer 1

Initial concentration of H₂ and F₂ will be 0.045/1.5 = 0.03 M

$4.3 = \frac{(2x)^2}{(0.03-x)(0.03-x)} \\ \sqrt{4.3} = \frac{2x}{0.03-x} \\ 2.073 = \frac{2x}{0.03-x} \\ 0.06219 -2.077x = 2x \\ 0.06219 = 4.077x \\ x = 0.015 \;M$

The equilibrium concentration of HF is $2 \times 0.015 = 0.03 \; M$

Answer: 0.03 M