Chemistry 3/4 2013 Thread

[Scooby]

I know that K1 is smaller than K2 but I don't really understand why... I mean I get that carbon monoxide has a greater affinity for haemoglobin than oxygen but how does that affect the extent of the reaction?

[09Ti08]

I know that K1 is smaller than K2 but I don't really understand why... I mean I get that carbon monoxide has a greater affinity for haemoglobin than oxygen but how does that affect the extent of the reaction?

I normally think about K as an indicator of the amount of products compared to the amount of reactants. What does that mean? That means that if you have a relatively big value for K, you have more products. If you are not convinced, think about the definition of K.

Now, in this question, they say that " carbon monoxide competes very successfully with oxygen for haemoglobin". This means that if the two react with haemoglobin, the amount of product created by the reaction by the reaction between carbon monoxide and haemoglobin is more than that between oxygen and haemoglobin. Therefore, K1<K2.

[09Ti08]

You mean the mol bit?
What's a mole? A mole is a number that is equal to the number of carbon-12 atoms that make up 12 grams. In short, it is roughly equal to 6.02*10^23 units.
So a mole of reaction means 6.02*10^23 copies of this reaction.
So with your reaction, if 1.204*10^24 magnesium atoms and 6.02*10^23 oxygen atoms reacted together to form 1.204*10^24 magnesium oxide units (better term for fundamental particles of ionic compounds?), you would have one mole of reaction and hence an enthalpy decrease of 1200 kJ

So your last statement is correct; unit refers to the equation as a whole. That's why if you double the coefficients, the enthalpy change doubles too.

Oh yes, but it's still a bit confusing. I normally leave the mol bit out and just calculate things normally using the balanced equation. Umm, the unit for energy is normally kJ in Physics.

[Scooby]

"When hydrogen iodide is placed in a closed container it will decompose into its elements hydrogen and iodine and an equilibrium will be established:

2HI(g) ↔ H₂(g) + I₂(g)

Initial concentration of HI: 0.025 mol/L

K_C=2.2x10

Calculate the equilibrium concentration for HI, H₂, and I₂."

How... 

[psyxwar]

"When hydrogen iodide is placed in a closed container it will decompose into its elements hydrogen and iodine and an equilibrium will be established:

2HI(g) ↔ H₂(g) + I₂(g)

Initial concentration of HI: 0.025 mol/L

K_C=2.2x10

Calculate the equilibrium concentration for HI, H₂, and I₂."

How... 

Kc = [H2]][I2]/[HI]^2 right?

[H2] = [I2] = x (equilbrium concentrations)

Use your ice box (initial, change, equilibrium) to come up with the equation for Kc in terms of x, then solve that.

[lzxnl]

"When hydrogen iodide is placed in a closed container it will decompose into its elements hydrogen and iodine and an equilibrium will be established:

2HI(g) ↔ H₂(g) + I₂(g)

Initial concentration of HI: 0.025 mol/L

K_C=2.2x10

Calculate the equilibrium concentration for HI, H₂, and I₂."

How... 

Let the concentration of HI that reacts be x. Yes, I mean concentration. Concentration ratios are mole ratios and vice versa.
This means that we'll have x/2 M of H2 forming and x/2 M of I2 forming.
And you'll have 0.025 - x M HI left
Now 22 = [H2][I2][HI]^-2
So 22 = x^2/4 * (0.025-x)^-2

Oh look, a quadratic equation! You can solve this.
Then double the x value to get the concentrations of H2 and I2

[thushan]

"When hydrogen iodide is placed in a closed container it will decompose into its elements hydrogen and iodine and an equilibrium will be established:

2HI(g) ↔ H₂(g) + I₂(g)

Initial concentration of HI: 0.025 mol/L

K_C=2.2x10

Calculate the equilibrium concentration for HI, H₂, and I₂."

How... 

Alternatively, you could assume that complete reaction has effectively occurred since Kc is so high (2.2 x 10^10), so [H2] = [I2] ~ 0.025 mol/L, then work out [HI] (whose value you cannot make such an assumption on).

Nliu95, good solution - but Year 12s aren't expected to know how to use quadratic equations to solve problems like this.

[Scooby]

Alternatively, you could assume that complete reaction has effectively occurred since Kc is so high (2.2 x 10^10), so [H2] = [I2] ~ 0.025 mol/L, then work out [HI] (whose value you cannot make such an assumption on).

Nliu95, good solution - but Year 12s aren't expected to know how to use quadratic equations to solve problems like this.

Woops, just realised the K value didn't copy across right. It's supposed to be K = 2.2x10^-3

[thushan]

Woops, just realised the K value didn't copy across right. It's supposed to be K = 2.2x10^-3

In that case you could get away with assuming that [HI]eq ~ 0.025 M.

[lzxnl]

Alternatively, you could assume that complete reaction has effectively occurred since Kc is so high (2.2 x 10^10), so [H2] = [I2] ~ 0.025 mol/L, then work out [HI] (whose value you cannot make such an assumption on).

Nliu95, good solution - but Year 12s aren't expected to know how to use quadratic equations to solve problems like this.

VCE Chem = year 12 chem
Quadratic equations = year 9 maths

Using quadratic equations too advanced for VCE chem. Logic?

[Jeggz]

Can someone please explain why DNA is acidic?
Thankyouu

[Alwin]

Can someone please explain why DNA is acidic?
Thankyouu

The DNA backbone contains phosphoric acid groups, which contribute H+ ions.

The phosphate ion is where the acidity is, its also called phosphoric acid while in your cells.
You could say that wouldn't it be equal with base and acidity because of the nitrogenous base, but the acid is so much powerful than the base, the base is there just basically for supporting the deoxyribose

Also, the "A" in DNA stands for acid

[Jeggz]

Why do Concentrated Chlorine ions always oxidise preferentially to water? :\

[Scooby]

When we say that the purpose of a monochromator in AAS, UV-Vis and IR is to "select the appropriate wavelength of EM radiation", what exactly does that mean?

Why in AAS is the EM radiation emitted from the hollow cathode lamp chopped into pulses?

Is AAS restricted to metal ions in solution or can other elements be analysed as well?

[Edward21]

When we say that the purpose of a monochromator in AAS, UV-Vis and IR is to "select the appropriate wavelength of EM radiation", what exactly does that mean?

Why in AAS is the EM radiation emitted from the hollow cathode lamp chopped into pulses?

Is AAS restricted to metal ions in solution or can other elements be analysed as well?

1) It is to select the wavelength most effectively absorbed by the sample, this will give the most accurate results, like when you do an absorbance graph of UV-Vis you choose the most absorbed wavelength because it will give the most accurate results.
2) It is chopped so the detecter can distinguish between the light being emitted from the lamp, and the atomiser/flame so it can accurately measure how much of the EM is being absorbed by the sample.
3) Metal ions, when in the atomiser, the solution is evaporated and the ions are atomised into, you guessed it, atoms! Everything else would most likely blow up or disintegrate, imagine ethanol in the flame.. BOOM!  it's a little common sense with knowing that other organics would combust, they're quantitatively analysed by gas chromatography anyway, or HPLC is they're too large to be vaporised, and gas chromatography is conducted at high temperatures, so they need to be vaporised, but not explode either!!