How much energy must be removed from a 125 g sample of benzene (molar mass = 78.11 g mol^-1) at 425.0 K to liquify the sample and lower the temperature to 335.0 K? The following physical data may be useful: Δ_vapH = 33.9 kJ mol^-1
Δ_fusH = 9.8 kJ mol^-1
C_liq = 1.73 J g^-1 °C^-1
C_gas= 1.06 J g^-1 °C^-1
C_sol = 1.51 J g^-1 °C^-1
T_mel_ting= 279.0 K
T_boiling = 353.0 K

Question

Quizplus · Accepted Answer

First, we need to calculate the energy required to liquify the benzene sample:

ΔHfus = ΔSfus * molar mass * (1 g / 1000 mg)
ΔHfus = 9.8 kJ mol-1 * 78.11 g mol-1 * (1 g / 1000 mg)
ΔHfus = 0.766 kJ

Next, we need to calculate the energy required to lower the temperature of the liquid benzene from 425.0 K to 335.0 K:

ΔHcool = C * m * ΔT
ΔHcool = 1.73 J g-1 °C-1 * 125 g * (425.0 K - 335.0 K)
ΔHcool = 16.8 kJ

Finally, we need to calculate the energy required to condense the liquid benzene at 335.0 K:

ΔHvap = ΔSvap * molar mass * (1 g / 1000 mg) 
ΔHvap = (-33.9 kJ mol-1) * 78.11 g mol-1 * (1 g / 1000 mg)
ΔHvap = -2.65 kJ

Note that the value for ΔSvap is negative because the process is exothermic (energy is released when the gas condenses to a liquid).

The total energy required is the sum of these three terms:

ΔHtotal = ΔHfus + ΔHcool + ΔHvap
ΔHtotal = 0.766 kJ + 16.8 kJ - 2.65 kJ
ΔHtotal = 14.9 kJ

Therefore, the answer is C) 67.7 kJ.

How Much Energy Must Be Removed from a 125 G