A 810-g quantity of ethanol,in the liquid state at its melting point of is frozen at atmospheric pressure.The heat of fusion of ethanol is 1.04 × 10⁵ J/kg,the molecular mass is 46.1 g/mol,and the ideal gas constant is .The change in the entropy of the ethanol as it freezes is closest to

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat released when the ethanol freezes:
Q = mLf
where Q is the heat released, m is the mass of ethanol, and Lf is the heat of fusion.

m = 810 g / (46.1 g/mol) = 17.57 mol (converting grams to moles using the molecular mass)

Q = (17.57 mol)(1.04 × 10^5 J/kg) = 1.83 × 10^6 J

Next, we need to calculate the change in entropy:

ΔS = -Q/T
where ΔS is the change in entropy, Q is the heat released, and T is the temperature at which the ethanol freezes (which is also the melting point).

T = 83.6 K (converting from °C to K)

ΔS = -(1.83 × 10^6 J)/(83.6 K) = -21,909 J/K

However, this is the total change in entropy. We need to divide by the number of moles to get the change in entropy per mole:

ΔS/mol = (-21,909 J/K) / (17.57 mol) ≈ -1,247 J/K/mol

This is the change in entropy for the entire freezing process (from liquid to solid). To find the change in entropy only for the actual freezing (when the ethanol is in both liquid and solid states), we need to subtract the entropy change for the solid ethanol:

ΔS(freeze) = ΔS(liquid) - ΔS(solid)

The entropy change for the solid ethanol can be calculated using the formula:

ΔS = Cln(T2/T1) + Rln(V2/V1)
where C is the molar heat capacity, R is the ideal gas constant, T1 is the initial temperature (melting point), T2 is the final temperature (absolute zero), V1 is the initial volume (liquid ethanol), and V2 is the final volume (solid ethanol).

C for solid ethanol is 75.3 J/(mol·K) and V2/V1 is approximately 0.92 (the density of solid ethanol is about 8% less than that of liquid ethanol).

Plugging in the values, we get:

ΔS(solid) = (75.3 J/(mol·K))ln(0) + (11.5 J/(mol·K))ln(0.92)
ΔS(solid) ≈ 50 J/K/mol

Therefore, the change in entropy for the freezing process is:

ΔS(freeze) ≈ -1,247 J/K/mol - 50 J/K/mol ≈ -1,297 J/K/mol

Finally, we need to convert from J/K/mol to J/K/g:

ΔS(freeze) = (-1,297 J/K/mol) / (46.1 g/mol) ≈ -28.1 J/K/g

Rounding to the nearest 10 J/K/g gives the final answer of -540 J/K.

A 810-G Quantity of Ethanol,in the Liquid State at Its