What is the concentration of HF under these conditions (listed in details)?

For the reaction:
H2(g) + F2(g) ⇆ 2HF(g)
K = 2.1 × 10-3 at a certain temperature.
At equilibrium: [H2] = [F2] = 0.066 M

1 Answer
Apr 9, 2018

#[HF]=0.03M#

Explanation:

value of #K_c=2.1*10^-3# at temp. constt.
At equilibrium: [H2] = [F2] = 0.066 M
so, #K_c=[HF]^2/([H_2][F_2])#
#2.1*10^-3=[HF]^2/(0.066*0.066)#
#[HF]^2=2.1*10^-3*(0.066)^2#
#[HF]^2=0.091476*10^-2=0.30/10=0.03M#