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3 years ago

What is the entropy change of the surroundings

Chemistry

1 answer:

KiRa [710]3 years ago

5 0

Answer: The entropy change of the surroundings will be -17.7 J/K mol.

Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol

Amount of Acetone given = 10.8 g

Number of moles is calculated by using the formula:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of acetone = 58 g/mol

Number of moles = $\frac{10.8}{58}=0.1862moles$

If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then

0.1862 moles will have = $\frac{32.3}{1}\times 0.1862=5.828kJ/mol$

To calculate the entropy change for the system, we use the formula:

$\Delta S_{sys}=\frac{\Delta H_{vap}}{T(\text{ in K)}}$

Temperature = 56.2°C = (273 + 56.2)K = 329.2K

Putting values in above equation, we get

$\Delta S_{sys}=\frac{5.828}{329.2}=0.0177kJ/Kmol=17.7J/Kmol$ (Conversion Factor: 1 kJ = 1000J)

At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,

$\Delta S_{system}+\Delta S_{surrounding}=0$

$\Delta S_{surrouding}=-\Delta S_{system}=-17.7J/Kmol$

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In a chemical reaction, exactly 2 mol of substance A react to produce exactly 3 mol of substance B. 2 A ⟶ 3 B How many molecules

Harman [31]

Answer:

There will be produced 1.71 moles of B which contain 1.03×10²⁴ molecules

Explanation:

The example reaction is:

2A → 3B

2 moles of A produce 3 moles of B

If we have the mass of A, we convert it to moles and then, we make the rule of three: 29.2 g / 25.6g/mol = 1.14 moles

Therefore 2 moles of A produce 3 moles of B

1.14 moles of A will produce (1.14 . 3) / 2 = 1.71 moles of B are produced

Now we can determine, the number of molecules

1 mol has NA molecules (6.02×10²³)

1.71 moles have (1.71 . NA) = 1.03×10²⁴ molecules

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2 years ago

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$\large \boxed{\text{4 600 000 kg or 4600 Mg}}$

Explanation:

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