ZuoTi.Pro
Question

For the reaction below at a certain temperature, it is found that the equilibrium concentrations in...

For the reaction below at a certain temperature, it is found that the equilibrium concentrations in a 5.05-L rigid container are [H2] = 0.0523 M, [F2] = 0.0121 M, and [HF] = 0.450 M.

H2(g) + F2(g) equilibrium reaction arrow 2 HF(g)

If 0.203 mol of F2 is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

science   chemistry   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Answer #1

Concentration equilibrium constant, Kc = [HF]eq2 / [H2]eq[F2]eq

Kc = (0.450 M)2 / [(0.0523 M) * (0.0121 M)2]

Kc = 320

moles F2 added = 0.203 mol

concentration F2 added = (moles F2 added) / (volume of container)

concentration F2 added = (0.203 mol) / (5.05 L)

concentration F2 added = 0.0402 M

new concentration F2 = (0.0121 M) + (0.0402 M)

new concentration F2 = 0.0523 M

ICE table H2 (g) F2 (g) \rightleftharpoons 2 HF (g)
Initial conc. 0.0523 M 0.0523 M 0.450 M
Change -x -x +2x
Equilibrium conc. 0.0523 M - x 0.0523 M - x 0.450 M + 2x

Kc = [HF]eq2 / [H2]eq[F2]eq

320 = (0.450 M + 2x)2 / [(0.0523 M - x) * (0.0523 M - x)]

Solving for x, x = 0.0244 M

equilibrium concentration H2 = 0.0523 M - x

equilibrium concentration H2 = 0.0523 M - 0.0244 M

equilibrium concentration H2 = 00279 M

equilibrium concentration F2 = 0.0523 M - x

equilibrium concentration F2 = 0.0523 M - 0.0244 M

equilibrium concentration F2 = 00279 M

equilibrium concentration HF = 0.450 M + 2x

equilibrium concentration HF = 0.450 M + 2 * (0.0244 M)

equilibrium concentration HF = 0.499 M

0 0
Add a comment
Know the answer?
Add Answer to:
For the reaction below at a certain temperature, it is found that the equilibrium concentrations in...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions

  • 1) The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g)...

    1) The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.311 M HI, 4.18×10-2 M H2 and 4.18×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.85×10-2 mol of I2(g) is added to the flask? 2) The equilibrium constant, K, for the following reaction is 1.20×10-2 at...

  • The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) +...

    The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.302 M HI, 4.05×10-2 M H2 and 4.05×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.203 mol of HI(g) is added to the flask?

  • Equilibrium Concentrations -- A + B = 2C At a particular temperature, K = 1.00×102 for...

    Equilibrium Concentrations -- A + B = 2C At a particular temperature, K = 1.00×102 for the reaction: H2(g) + F2(g) 2HF(g) In an experiment, at this temperature, 1.00×10-1 mol of H2 and 1.00×10-1 mol of F2 are introduced into a 1.09-L flask and allowed to react. At equilibrium, all species remain in the gas phase. What is the equilibrium concentration (in mol/L) of H2? mol/L 1 pts What is the equilibrium concentration (in mol/L) of HF? mol/L 1 pts...

  • 6) Distrubuting Equilibrium a) The equilibrium constant, K, for the following reaction is 10.5 at...

    6) Distrubuting Equilibrium a) The equilibrium constant, K, for the following reaction is 10.5 at 350 K 2CH2Cl2(g)CH4(g + CCl4(g) An equilibrium mixture of the three gases in a 1.00 L flask at 350 K contains 5.32x10-2 M CH2C2, 0.172 M CH4 and 0.172 M CCI4. What will be the concentrations of the three gases once equilibrium has been reestablished, if 9.58x102 mol of CH4(g is added to the flask? CH2Cl2l [CH4 [CCI4] b) The equilibrium constant, K, for the...

  • The equilibrium constant, Kc, for the following reaction is 55.6 at 698 K: H2(g) + I2(g)...

    The equilibrium constant, Kc, for the following reaction is 55.6 at 698 K: H2(g) + I2(g) ---------->2HI(g) 1) Calculate the equilibrium concentrations of reactants and product when 0.309 moles of H2 and 0.309 moles of I2 are introduced into a 1.00 L vessel at 698 K. [H2] = M? [I2] = M? [HI] = M? 2.The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K: PCl5(g)------->PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a...

  • The equilibrium constant, K, for the following reaction is 0.186 at 332 K. NH4HS(s) (equilibrium) NH3(g)...

    The equilibrium constant, K, for the following reaction is 0.186 at 332 K. NH4HS(s) (equilibrium) NH3(g) + H2S(g) An equilibrium mixture in a 16.2 L container at 332 K contains 0.322 mol  NH4HS(s), 0.543 M NH3 and 0.343 M H2S. What will be the concentrations of the two gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 8.20 L? [NH3] = M [H2S] = M

  • The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) +...

    The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.329 M HI, 4.41×10-2 M H2 and 4.41×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 2.54×10-2 mol of H2(g) is added to the flask? [HI] = M [H2] = M [I2] = M

  • The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) +...

    The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.319 M HI, 4.27×10-2 M H2 & 4.27×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.224 mol of HI(g) is added to the flask? [HI] = M [H2] = M [I2] = M please help me!

  • The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) +...

    The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.316 M HI, 4.24×10-2 M H2 and 4.24×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.21×10-2 mol of I2(g) is added to the flask?

  • The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) ----> H2(g)...

    The equilibrium constant, K, for the following reaction is 1.80×10-2 at 698 K. 2HI(g) ----> H2(g) + I2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 698 K contains 0.306 M HI, 4.10×10-2 M H2 and 4.10×10-2 M I2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 0.208 mol of HI(g) is added to the flask? [HI]   = ______ M [H2]   = ______ M [I2]   = ______M

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT