At a particular temperature, K = 1.00×10^{2} for the reaction:

What is the new equilbrium concentration (in mol/L) of HF?

Initial concentrations: H2: 1.9mol/1.25L=1.52mol/L F2: 1.9mol/1.25L=1.52mol/L

Using ICE Table: H2 F2 2HF

I: 1.52 1.52 0

C: x x 2x

E: 1.52-x 1.52-x 2x

Since K= Product/ reactant

100=(2x)^2/ (1.52-x) ^2

To solve for x take square root of both sides: 10=2x/(1.52-x)

15.2-10x=2x

15.2=12x

x=1.27

The final concentration of H2 is 1.52-x=1.51-1.27=0.25

0.25mol/L is the answer

adding 0.95: H2 is now 1.2mol/L now F2 is 0.25mol/L and is limiting

so all F2 react giving 0.25*2 more HF=0.5

adding 2x form before 2.54+0.5 =3.04 :)

Please rate my answer best and give a 5 star ratings. You donot need to pay for besting answers ;)

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up