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musickatia [10]
3 years ago
11

Which relationship or statement best describes delta s for the following reaction:

Chemistry
1 answer:
zalisa [80]3 years ago
4 0

Answer:

b) Delta S < 0

Explanation:

The change in the entropy (ΔS) is related to the change in the number of gaseous moles of the reaction: Δn(g) = n(g, products) - n(g, reactants).

  • If Δn(g) > 0, the entropy increases (ΔS > 0).
  • If Δn(g) < 0, the entropy decreases (ΔS < 0).
  • If Δn(g) = 0, there is little or no change in the entropy

Let's consider the following equation.

2 H₂S(g) + 3 O₂(g) → 2 H₂O(g)

Δn(g) = 2 - 5 = - 3. Since Δn(g) < 0, the entropy decreases and ΔS < 0.

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Pure chlorobenzene (C6H5Cl) has a normal boiling point of 131.00 °C. A solution of 32.5 g of 2,8-dibromodibenzofuran (C12H6Br2O)
vichka [17]

Answer:

Kb →  1.56 °C / m

Explanation:

This is all about boiling point elevation, the colligative property that shows that boiling point for a solution is higher than boiling point of pure solvent.

This is the formula: ΔT = Kb . m . i

where i is the Van't Hoff factor (ions dissolved in solution). As these are organic compounds, we assume they are non electrolytic,

m is molality (mol of solute / 1kg of solvent)

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ΔT = T° boiling from solution - T° boiling from solute

First of all, let's determine the moles of solute.

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Molality = 0.286 mol / 0.195 kg = 1.47 m

Let's replace the values in the formula

133.30 °C - 131°C = Kb . 1.47m .1

2.30°C / 1.47 m =  Kb →  1.56 °C / m

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