I'm stuck on how to do these 4 problems concerning Chemical Equilibriums.
When varsol and iodine/pottasium iodide solutions are mixed, the two phases form.. phase A (top) would be pink and phase B (bottom) would be brown.
What are the colours in each phases for these 4 additions to test tube control (described above)?
b) add 4 ml water
c) 2 drops saturated iodine in KI solution
d) 2 drops saturated sodium thiosulphate
e) 2 drops saturated sodium potassium iodide solution
An explanation would be much helpful. :)
Without knowing the concn of the original varsol layer and the original water layer, I can make only an educated guess but my take is as follows. I've never done this experiment so I'm making an educated guess.
b. Adding 4 mL water should not change the varsol layer but will dilute the water layer so it should become a lighter brown color.
c. Adding saturated I2 in KI solution I think will increase the concn of I2 in the varsol as well as increasing the concn of KI3 in the water. I think both pink/violet and brown colors will become more intense (darker pink/violet and darker brown).
d. Adding Na2S2O3 will react with the I2 as follows:
S2O3^2- + I3^- ==> S2O4^2- + 3I^-
I think the brown color will become lighter in the water solution because the I2/I3^- is being used. Since the distribution coefficient between varsol and water favors varsol, I don't think the pink/violet color in the varsol will be affected.
e. I'll leave this one for you. It is unclear to me if this is NaI + KI or some kind of double salt.
I don't understand the question. I understand that the varsol will be the top layer and it dissolves the I2 in a non-polar solvent so the color is pink. The lower level is water and it dissolves the KI/I2 mixture of KI3 to form a brown color. The brown color forms because the KI3 is dissolved in a polar solvent and that distorts the pink color to a brown color. What I don't understand is the experimental part below the introductory paragraph.
The experimental part is that, say we add 4 ml of water, 2 drops of saturated iodine in KI solution.. etc. to the reaction in the introductory paragraph. We are suppose to predict the colour changes in the two phases when each is added to the initial reaction.
Consider the equilibrium system: N204 (g) = 2 NO2 (g) for which the Kp = 0.1134 at 25 C and deltaH rx is 58.03 kJ/mol. Assume that 1 mole of N2O4 and 2 moles of NO2 are introduced into a 5 L contains. What will be the equilibrium value of [N204]?
Options are:
A) 0.358 M
B) 0.042 M
C) 0.0822 M
D) 0.928 M
E) 0.379 M
Please provide an explanation, I want to understand this question.. thank you so much!!
sorry.. accidentally posted on your question :/
thanks
To determine the colors in each phase for the four additions, we need to understand the chemical reactions that occur during each step.
Before we start, let's identify the colors of the original solutions:
- Varsol: Unknown color (control)
- Iodine/Potassium Iodide: Unknown color (control)
Now, let's go through each addition:
b) Add 4 ml of water:
Water does not typically react with Varsol or iodine/potassium iodide solutions. Therefore, the colors in each phase will remain the same as the original control.
- Phase A (top): Pink
- Phase B (bottom): Brown
c) Add 2 drops of saturated iodine in KI solution:
When iodine reacts with potassium iodide, a brown complex called triiodide ion (I3-) is formed. This reaction indicates the presence of iodine.
- Phase A (top): Pink (no change)
- Phase B (bottom): Brown
d) Add 2 drops of saturated sodium thiosulphate:
Sodium thiosulphate reacts with iodine to form sodium iodide and sodium tetrathionate. The iodine is consumed in this reaction, resulting in a color change.
- Phase A (top): Pink (no change)
- Phase B (bottom): Colorless (due to the consumption of iodine)
e) Add 2 drops of saturated sodium potassium iodide solution:
Adding sodium potassium iodide solution will introduce additional iodide ions into the system. In the presence of iodine, these iodide ions will react to form triiodide ion, resulting in a brown color.
- Phase A (top): Pink (no change)
- Phase B (bottom): Brown (due to the formation of triiodide ion)
So, to summarize:
b) Water does not cause a color change.
c) The addition of iodine in KI solution turns Phase B brown.
d) The addition of sodium thiosulphate makes Phase B colorless.
e) The addition of sodium potassium iodide solution turns Phase B brown.
Keep in mind that the colors may be subject to interpretation and variations based on concentrations and the specific nature of the chemicals used in the experiment.