please help within five minutes

label Chemistry
account_circle Unassigned
schedule 1 Day
account_balance_wallet $5

At a particular temperature, K = 1.00×102 for the reaction:

H2(g) + F2(g) 2HF(g)

To the mixture above, an additional 9.50×10-1 mol of H2 is added.

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

Feb 2nd, 2015

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)




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 ;)

Feb 2nd, 2015

Did you know? You can earn $20 for every friend you invite to Studypool!
Click here to
Refer a Friend
Feb 2nd, 2015
Feb 2nd, 2015
Oct 18th, 2017
Mark as Final Answer
Unmark as Final Answer
Final Answer

Secure Information

Content will be erased after question is completed.

Final Answer