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
Shubh
Apr 9, 2018

[HF]=0.03M[HF]=0.03M

Explanation:

value of K_c=2.1*10^-3Kc=2.1⋅10−3 at temp. constt.
At equilibrium: [H2] = [F2] = 0.066 M
so, K_c=[HF]^2/([H_2][F_2])Kc=[HF]2[H2][F2]
2.1*10^-3=[HF]^2/(0.066*0.066)2.1⋅10−3=[HF]20.066⋅0.066
[HF]^2=2.1*10^-3*(0.066)^2[HF]2=2.1⋅10−3⋅(0.066)2
[HF]^2=0.091476*10^-2=0.30/10=0.03M[HF]2=0.091476⋅10−2=0.3010=0.03M

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