hard industrial chem question help (1 Viewer)

[dongtamfp]

Hydrogen gas and iodine vapour react to form hydrogen iodide gas.
a) In one experiment, 4.00mol of hydrogen iodide gas was added to a 2 litre reaction vessel at an elevated temperature. THe system was allowed to reach equilibrium and the concentration of hydrogen iodide remaining was measured and foudn to be 3.10 mol. Determine the equilibrium constant for this reaction.
b) Once equilibrium is established, an additional 0.3 mol of HI is injected into the vessel at constant temperature.
i) Demonstrate that the system is no longer at equilibrium
ii) Verify that the system will attain equilibrium once more when the concentration of each species reach the following (approximate) values:
[HI] = 1.666 mol/L
[H2] = [I2]
= 0.242 mol/L

i have no idea how to do this guys.

 

[MaccaFacta]

The reaction is 2HI(g) = H2(g) + I2(g).

If 4.00 moles of HI are put into a 2 L vessel and, once the above equilibrium has been established, 3.10 moles remain, then 0.90 moles of HI must have reacted with itself to produce 0.45 moles of H2 and 0.45 moles of I2. Imagine 2 HI molecules smashing into each other forming one molecule of H2 and one molecule of I2.

Now that we have the number of moles of each of the 3 substances we can calculate their concentrations in the 2L vessel. We have 3.10 moles of HI, 0.45 moles of H2 and 0.45 moles of I2.

Therefore the [HI] = 3.10 moles / 2 L = 1.55 mol / L, [H2] = 0.45 / 2 = 0.225 mol/L and [I2] also equals 0.45 mol/L.

Now substitute into the equilibrium constant expression K = [H2][I2]/[HI]^2 = 0.45*0.45/1.55^2 = 0.0843

I gotta go. I'll come back for the rest of this question tomorrow...

 

[Thank_You]

For part a) you always do that 'table' method or w/e it's called. I always do it, and I have no problems. Example of this, if you get your hands on any CSSA papers, and check solutions for these type of questions, they always set it up as initial conc., change in conc., and equilibrium conc.. Then you just apply the equilibrium constant formula

For part b i) - I THINK you just, from the table, just add 0.3mol to the equilibrium value of HI. Apply the quotient formula, and you will find Q does not equal K. Hence not equilibrium
ii) Use those given values, and apply equilibrium constant again...

I hope @maccafacta can answer your questions cos i cbf answering lol, busy with studying english trials !

 

[cub3root]

For part a) you always do that 'table' method or w/e it's called. I always do it, and I have no problems. Example of this, if you get your hands on any CSSA papers, and check solutions for these type of questions, they always set it up as initial conc., change in conc., and equilibrium conc.. Then you just apply the equilibrium constant formula

For part b i) - I THINK you just, from the table, just add 0.3mol to the equilibrium value of HI. Apply the quotient formula, and you will find Q does not equal K. Hence not equilibrium
ii) Use those given values, and apply equilibrium constant again...

I hope @maccafacta can answer your questions cos i cbf answering lol, busy with studying english trials !

then why are you on BoS?

 

[Thank_You]

^procrastination

 

[cub3root]

lel

 

[Crisium]

then why are you on BoS?

The question you should be asking:

Why are they answering a question that was asked a year ago?