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artcher [175]
2 years ago
9

The heat of vaporization of acetic acid is . Calculate the change in entropy when of acetic acid condenses at .

Chemistry
1 answer:
Umnica [9.8K]2 years ago
8 0

The question is incomplete, the complete question is;

The heat of vaporization ΔHv of acetic acid HCH3CO2 is 41.0 /kJmol. Calculate the change in entropy ΔS when 954.g of acetic acid condenses at 118.1°C. Be sure your answer contains a unit symbol. Round your answer to 3 significant digits.

Answer:

-1.67 JK-1

Explanation:

Since Heat of vaporization of acetic acid = 41.0 kJ/mol

Therefore:

Heat of condensation of acetic acid = -41.0 kJ/mol

Mass of acetic acid = 954 g

Temperature of condensation = 118.1 °C or 391.1 K

Number of moles of acetic acid = 954 g/60g/mol = 15.9 moles

Heat evolved during condensation = 15.9 moles *  -41.0 kJ/mol = -651.9 KJ

Entropy change (ΔS) = Heat evolved/ Temperature = -651.9 KJ/391.1 K

Entropy change (ΔS) = -1.67 JK-1

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A bomb calorimeter has a heat capacity of 783 J/oC and contains 254 g of water whose specific heat capacity is 4.184 J/goC. How
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Answer : The amount of heat evolved by a reaction is, 4.81 kJ

Explanation :

Heat released by the reaction = Heat absorbed by the calorimeter + Heat absorbed by the water

q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

where,

q = heat released by the reaction

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

c_1 = specific heat of calorimeter = 783J/^oC

c_2 = specific heat of water = 4.184J/g^oC

m_2 = mass of water = 254 g

\Delta T = change in temperature = T_2-T_1=(23.73-26.01)=-2.28^oC

Now put all the given values in the above formula, we get:

q=[(783J/^oC\times -2.28^oC)+(254g\times 4.184J/g^oC\times -2.28^oC)]

q=-4208.28J=-4.81kJ

Therefore, the amount of heat evolved by a reaction is, 4.81 kJ

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