Calculate the amount of enthalpy required to heat 25.0 g of solid benzene (C₆H₆) at -10^$\circ$C to liquid benzene at 20.0^$\circ$C. Thermodynamic data for benzene: specific heat of solid benzene = 1.52 J/g·^$\circ$C; specific heat of liquid benzene = 1.73 J/g·^$\circ$C; melting point = 5.5^$\circ$C; $\Delta$H_fus = 9.9 kJ/mol. A) 3.8 kJ B) 4.1 kJ C) 4.4 kJ D) 4.7 kJ E) None of the above

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of heat required to heat the solid benzene from -10°C to its melting point of 5.5°C:

q = mCΔT
q = (25.0 g)(1.52 J/g·°C)(5.5°C - (-10°C))
q = 2,990 J

Next, we need to calculate the amount of heat required to melt the benzene:

q = nΔHfus
n = m/Mw = 25.0 g / 78.11 g/mol = 0.3207 mol
q = (0.3207 mol)(9.9 kJ/mol)
q = 3.18 kJ

Finally, we need to calculate the amount of heat required to heat the liquid benzene from its melting point of 5.5°C to 20.0°C:

q = mCΔT
q = (25.0 g)(1.73 J/g·°C)(20.0°C - 5.5°C)
q = 646 J

The total amount of heat required is the sum of these three values:

qtotal = 2,990 J + 3.18 kJ + 646 J
qtotal = 4.46 kJ

The closest option is C, 4.4 kJ, so that is our answer.

Calculate the Amount of Enthalpy Required to Heat 25 ∘\circ∘ C to Liquid Benzene at 20

Calculate the Amount of Enthalpy Required to Heat 25 $\circ$ C to Liquid Benzene at 20