2007 Questions

[Aaron.Judd]

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Q13: Why is the answer A?

Q22 a) How do you do this?

Q22 b) ii) Whats this answer?

Q 27 a/b) How do you work this out, especially the calculation?

[Aaron.Judd]

Nevermind about 22, a I get that now.

[JasonNg1025]

13) Either by reasoning that the other ones are inaccurate (bad way of doing), or reasoning:

N_{2 (g)} + 2H_{2 (g)} <=> 2NH_{3 (g)} + heat (exothermic)

There are clearly more moles of gas on the R.H.S. Decreasing in pressure, therefore, shifts equilibrium to the left. Since the forward reaction releases heat, the reverse reaction absorbs heat, hence the answer is A, absorbing heat.

22) a) Look at your other thread

22) b) ii) The sulfur reduction policy should lead to lower levels of SO₂, as:

S + O₂ --> SO₂.

SO₂ forms acid rain:

SO₂ + H₂O --> H₂SO₃

2SO₂ + O₂ --> 2SO₃

SO₃ + H₂O --> H₂SO₄

Acid rain reduces pH of water bodies, killing less tolerant aquatic life. It reduces plant growth and even kills some plants. Acid rain also corrodes the protective layer on leaves, making them more vulnerable and less tolerant. Limestone buildings and monuments can also be corroded.

The reduce in SO₂ also increases general air quality. Therefore, the sulfur reduction policy will be good for the environment.

27) a) Basically (elaborate on this), use scales to weigh water sample, add AgNO₃ in excess to fully precipitate out Cl^-, pour the reacted water sample through the sintered glass filter. The residue should be AgCl precipitate. Dry, and weigh. Use this to calculate Cl^- concentration.

27) b) Precipitate is 3.65g, => moles AgCl = 0.0255.

AgCl --> Ag⁺ + Cl^- => moles Cl^- = 0.0255.

Mass Cl^- = 0.903g

This is in 50mL (we assume 50g) of water.

Concentration in ppm = 0.903 / 50 x 1000000ppm

This is 18052.67 ppm

[Aaron.Judd]

With 13
Initally there is 4 mols of gas on the left to 2 mols of gas on the right. So if you reduce the entire system pressure, I.e. half it. Then you have 2 mols on left to 1 mol of gas on the right. Hence shouldn't the reaction still be reactants to products (i.e. to the right)?

[Aaron.Judd]

What about 24 b) ii)

Thanks for all your help so far.

[JasonNg1025]

Originally Posted by Aaron.Judd

With 13
Initally there is 4 mols of gas on the left to 2 mols of gas on the right. So if you reduce the entire system pressure, I.e. half it. Then you have 2 mols on left to 1 mol of gas on the right. Hence shouldn't the reaction still be reactants to products (i.e. to the right)?

When you reduce the pressure on the system, that's what initially happens. Then the system wants to make up for the pressure it lost, hence it will favour the side with more moles. Hope that makes sense...

Originally Posted by Aaron.Judd

What about 24 b) ii)

Conditions are different - the theoretical value uses standard conditions (25^o C and 100kPa I think), which may differ it a little. Also, in practice, most of the time you get incomplete combustion, making your empirical value much lower.

[danz90]

Did you get like 2.3L for 22(a) ??

[Aaron.Judd]

Originally Posted by danz90

Did you get like 2.3L for 22(a) ??

Yes

[Aaron.Judd]

Thanks for the help with 27.

[Marinatos]

Since you worked out how to do it, how do you do Q22. I'm so brain dead right now.

edit: nver mind. I did eventually get it.

[:::a:::]

omg nevermind (just realised)