Equilibrium between [math]CH_3OH[/math] and [math]CH_3I, OH^-[/math]

[Dhamnekar Win,odd]

 I am working on question (d) for answering it. Any chemistry help will be accepted. Do you know the correct answer for question(d)? 

Edited April 1, 2022 by Dhamnekar Win,odd

[exchemist]

1 hour ago, Dhamnekar Win,odd said:

 I am working on question (d) for answering it. Any chemistry help will be accepted. Do you know the correct answer for question(d)? 

You've got in a slight muddle with your equation and the equilibrium constant derived from it. What you have been given is the equilibrium constant for the reaction you were given, not the new one you incorrectly generated on the basis of molecular iodine. You do not have molecular iodine here. You have iodide ions in solution. To balance the reaction there will be some cation, e.g. K+, which does not participate but features on both sides. Apart from the fact that the charges don't balance in the reaction you generated, you can't take an an equilibrium constant you have been given for one reaction and start applying it to a different reaction!  

I think that, as a result, you have ended up with a factor of 4 in the numerator that should not be there. I did it without that factor and got 2 roots to the quadratic, one of which is 0.38 (the other being 0.73 which is clearly nonsense in this context). 

With a bit of luck once you have resolved this the rest will come out.