allthebest Posted June 21, 2016 Report Share Posted June 21, 2016 (edited) 0.50 mol of I2(g) and 0.50 mol of Br2(g) are placed in a closed flask. The following equilibrium is established. I2(g) + Br2(g) IBr(g) The equilibrium mixture contains 0.80 mol of IBr(g). What is the value of Kc? How is the answer 64? Should I change the mol into mol dm-3? Thank you. Edited June 22, 2016 by allthebest Reply Link to post Share on other sites More sharing options...
kw0573 Posted June 22, 2016 Report Share Posted June 22, 2016 (edited) Hints: 1) Balance the equation 2) Assume a variable V for volume of container and calculate concentration using V. 3) What are the concentration of reactants at equilibrium? (Why do you need this?) Plug into definition of equilibrium constant and simplify! PM me if you are still stuck. Edited June 22, 2016 by kw0573 Reply Link to post Share on other sites More sharing options...
allthebest Posted June 22, 2016 Author Report Share Posted June 22, 2016 2 hours ago, kw0573 said: In the interest of providing you an answer. You should give us the COMPLETE question. You didn't even give ANY question. When you update the question I'll update this answer. Sorry, I forgot to put the question. I've updated now. Thanks 1 Reply Link to post Share on other sites More sharing options...
Msj Chem Posted June 23, 2016 Report Share Posted June 23, 2016 0.8 mol of IBr reacts with 0.4 mol of I2 and Br2 (because of the 2:1 ratio). So the equilibrium mixture has 0.8 mol of IBr and 0.1 mol of I2 and Br2. Write the equilibrium expression, plug in those equilibrium values and you'll get 64 (0.64/0.01). Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.