Question

1) Some C₆H₅CH₂OH is allowed to dissociate into C₆H₅CHO and H₂ at 523 K. At equilibrium, [C₆H₅CH₂OH] = 0.212 M, and [C₆H₅CHO] = [H₂] = 5.26×10^-2 M. Additional C₆H₅CH₂OH is added so that [C₆H₅CH₂OH]_new = 0.362 M and the system is allowed to once again reach equilibrium.

C₆H₅CH₂OH(g) --> C₆H₅CHO(g) + H₂(g) K = 1.30×10^-2 at 523 K

(a) In which direction will the reaction proceed to reach equilibrium? _________

to the right? to the left?

(b) What are the new concentrations of reactants and products after the system reaches equilibrium?

[C6H5CH2OH]	=__	M
[C6H5CHO]	= __	M
[H2]	= __	M

2) Consider the equilibrium between (CH₃)₂CHOH, (CH₃)₂CO and H₂.

(CH₃)₂CHOH(g) ---> (CH₃)₂CO(g) + H₂(g) K = 3.43×10^-2 at 484 K

The reaction is allowed to reach equilibrium in a 7.50-L flask. At equilibrium, [(CH₃)₂CHOH] = 0.312 M, [(CH₃)₂CO] = 0.104 M and [H₂] = 0.104 M.

(a) The equilibrium mixture is transferred to a 15.0-L flask. In which direction will the reaction proceed to reach equilibrium?

_________. to the right? to the left?

(b) Calculate the new equilibrium concentrations that result when the equilibrium mixture is transferred to a 15.0-L flask.

[(CH3)2CHOH]	=____	M
[(CH3)2CO]	= ___	M
[H2]	= ___	M

C) The equilibrium constant, K, for the following reaction is 1.29×10^-2 at 600 K.

COCl₂(g) --> CO(g) + Cl₂(g)

An equilibrium mixture of the three gases in a 1.00 L flask at 600 K contains  0.333 M  COCl₂, 6.55×10^-2 M CO and 6.55×10^-2 M Cl₂. What will be the concentrations of the three gases once equilibrium has been reestablished, if 4.75×10^-2mol of CO(g) is added to the flask?

[COCl2]	=	M
[CO]	=	M
[Cl2]	=	M

This is a problem I am stuck on. It's mostly because I don't know how to do the quadratic equation since the video skips the process. Can you please include that?

Answer 1

1.a) Right

1.b)

  [C₆H₅CH₂OH] = .333

  [(CH₃)₂CO] = .2935 = [H₂]

2.a) Right