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Scilla [17]
3 years ago
12

Consider two processes: sublimation of I2(s) and melting of I2(s) (Note: the latter process can occur at the same temperature bu

t somewhat higher pressure).
I2(s) → I2(g)
I2(s) → I2(l)
Is ΔS positive or negative in these processes?

Chemistry
2 answers:
AVprozaik [17]3 years ago
4 0

Answer:

positive

positive

Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by amount heat per change in temperature.

  • When solid is converted to gas entropy increases,

As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.

So,

The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,

And

If the particles are loosely held , the entropy will increase , i.e. , positive entropy .

Similar with solid converting to liquid , the entropy will increases , i.e. , positive entropy .

Hence ,

The correct sign of entropy for both the process is positive .

ziro4ka [17]3 years ago
3 0

Answer:

ΔS increases in each case.  

Explanation:

Entropy (S) is a measure of the degree of the spreading of energy within a system. The more ways that energy can be distributed, the greater the entropy.

(a) Sublimation of I₂

I₂(s)⇌ I₂(g)

In the solid, the I₂ molecules are locked in position in a crystal lattice. They can vibrate only slightly about their equilibrium locations.  

Those in the gas phase can move in different directions and at different speeds, so the thermal energy is widely distributed among many microstates. The mobility of the molecules is greatly increased, so the number of number of microstates increases and entropy increases.

(b) Melting of I₂

I₂(s)⇌ I₂(ℓ)

The molecules in the solid can only vibrate about a fixed position.

Those in the liquid phase can move around and slide past each other. The increased freedom of motion means that thermal energy is distributed among more microstates, so the entropy increases.

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0.4444 g/cm³ ≅ 0.44 g/cm³ (2 significant figures).

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<em>d = m/V,</em>

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Twenty five grams of Iron 3 oxide react with an excess of carbon monoxide to form 15 g of Fe. Carbon dioxide is the other produc
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<h3>Answer:</h3>

Theoretical mass = 17.42 g

Percent yield of Fe = 86.11%

<h3>Explanation:</h3>

The equation for the reaction between iron (iii) oxide and carbon monoxide is given by;

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

We are required to calculate the theoretical yield and the percentage yield of Iron.

Step 1: Moles of iron (iii) oxide

Moles are given by dividing the mass of the compound by the molar mass.

Molar mass of Iron(iii) oxide = 159.69 g/mol

Moles of Iron(III) oxide = 25 g ÷ 159.69 g/mol

                                     = 0.156 moles

Step 2: Moles of Iron produced

From the equation 1 mole of Iron(iii) oxide reacts to produce 2 moles of Fe.

Therefore, the mole ratio of Fe₂O₃ to Fe is 1 : 2.

Thus, moles of Fe = Moles of Fe₂O₃ × 2

                              = 0.156 moles × 2

                              = 0.312 moles

Step 3: Theoretical mass of iron produced

To calculate the mass of iron we multiply the number of moles of iron with the relative atomic mass.

Relative atomic mass = 55.845

Mass of iron = 0.312 moles × 55.845

                    = 17.42 g

Step 4: Percent yield of iron

% yield = (Actual mass ÷ Theoretical mass)×100

            = (15 g ÷ 17.42 g) × 100 %

            = 86.11%

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