Consider the equilibrium between C6H5CH2OH, C6H5CHO and H2. C6H5CH2OH(g) C6H5CHO(g) + H2(g) K = 2.41×10-2 at 552 K The reaction is allowed to reach equilibrium in a 9.40-L flask. At equilibrium, [C6H5CH2OH] = 0.406 M, [C6H5CHO] = 9.89×10-2 M and [H2] = 9.89×10-2 M. (a) The equilibrium mixture is transferred to a 18.8-L flask. In which direction will the reaction proceed to reach equilibrium?
Consider the equilibrium between C6H5CH2OH, C6H5CHO and H2.
C6H5CH2OH(g) C6H5CHO(g) + H2(g) K = 2.41×10-2 at 552 K
The reaction is allowed to reach equilibrium in a 9.40-L flask. At equilibrium, [C6H5CH2OH] = 0.406 M, [C6H5CHO] = 9.89×10-2 M and [H2] = 9.89×10-2 M.
(a) The equilibrium mixture is transferred to a 18.8-L flask. In which direction will the reaction proceed to reach equilibrium?
(b) Calculate the new equilibrium concentrations that result when the equilibrium mixture is transferred to a 18.8-L flask.
[C6H5CH2OH]M =
[C6H5CHO] M =
[H2] M =
If your equilibrium values are correct, when you substitute them into the equilibrium expression you should reproduce the value of K given. please do this part last two times i asked they didn't and gave me the wrong answer
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 3 images
