The heat of fusion Δ, of benzene (C6H6) is 10.6 kJ/mol. Calculate the change in entropy AS when 2.3 g of benzene freezes at 56 °C Be sure your answer contains a unit symbol. Round your answer to 2 significant digits.

Answers

Answer 1

The change in entropy when 2.3 g of benzene freezes at 56 °C is 0.9 J/K.

To calculate the change in entropy (ΔS) when 2.3 g of benzene freezes, we need to use the equation:

ΔS = ΔH / T

where ΔH is the heat of fusion, and T is the freezing temperature.

First, we need to convert the mass of benzene from grams to moles. The molar mass of benzene is:

C6H6: 6(12.01 g/mol) + 6(1.01 g/mol) = 78.11 g/mol

2.3 g / 78.11 g/mol = 0.0295 mol

Next, we need to calculate the heat absorbed by 0.0295 mol of benzene during freezing:

ΔH = nΔHf = (0.0295 mol)(10.6 kJ/mol) = 0.3127 kJ

Finally, we can calculate the change in entropy:

ΔS = ΔH / T = 0.3127 kJ / (56 + 273) K = 0.0009 kJ/K

We can convert the units of kJ/K to J/K:

0.0009 kJ/K x 1000 J/kJ = 0.9 J/K

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Explanation:

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