How much enthalpy is necessary to heat 10.0 g of solid benzene (C₆H₆) at 0.0^$\circ$C to benzene vapor at 100^$\circ$C
$~~~~~~~~~~~~~~~~~~~~~\text { Data for Benzene }$
$\begin{array}{|l|l|}\hline \text { melting point } & 5.5^{\circ} \mathrm{C} \\\hline \text { boiling point } & 80.1^{\circ} \mathrm{C} \\\hline \text { specific heat of solid benzene } & 1.52 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} \\\hline \text { specific heat of liquid benzene } & 1.73 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} \\\hline \text { specific heat of benzene vapor } & 1.06 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} \\\hline \Delta \mathrm{H}_{\text {fus }} & 9.9 \mathrm{~kJ} / \mathrm{mol} \\\hline \Delta \mathrm{H}_{\text {vap }} & 30.8 \mathrm{~kJ} / \mathrm{mol} \\\hline\end{array}$

Question

How much enthalpy is necessary to heat 10.0 g of solid benzene (C₆H₆) at 0.0^$\circ$C to benzene vapor at 100^$\circ$C

~~~~~~~~~~~~~~~~~~~~~\text { Data for Benzene }

\begin{array}{|l|l|}\hline \text { melting point } & 5.5^{\circ} \mathrm{C} \\\hline \text { boiling point } & 80.1^{\circ} \mathrm{C} \\\hline \text { specific heat of solid benzene } & 1.52 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} \\\hline \text { specific heat of liquid benzene } & 1.73 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} \\\hline \text { specific heat of benzene vapor } & 1.06 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C} \\\hline \Delta \mathrm{H}_{\text {fus }} & 9.9 \mathrm{~kJ} / \mathrm{mol} \\\hline \Delta \mathrm{H}_{\text {vap }} & 30.8 \mathrm{~kJ} / \mathrm{mol} \\\hline\end{array}

Quizplus · Accepted Answer

To calculate the enthalpy required to heat solid benzene to benzene vapor, we need to use the following formula:

ΔH = n x ΔHvap + ΔHfus + n x Cp x ΔT

where ΔH is the enthalpy change, n is the number of moles, ΔHvap is the enthalpy of vaporization, ΔHfus is the enthalpy of fusion, Cp is the heat capacity, and ΔT is the change in temperature.

First, we need to calculate the number of moles of benzene:

molar mass of benzene (C6H6) = 12.01 x 6 + 1.01 x 6 = 78.11 g/mol
number of moles = 10.0 g / 78.11 g/mol = 0.128 moles

Next, we need to calculate the enthalpy of vaporization of benzene. According to the table of thermodynamic values, the enthalpy of vaporization of benzene is 30.8 kJ/mol.

ΔHvap = 0.128 moles x 30.8 kJ/mol = 3.94 kJ

The enthalpy of fusion of benzene is negligible, so we can skip that term.

Finally, we need to calculate the enthalpy change due to the change in temperature. The heat capacity of benzene is 0.138 J/g*K.

ΔT = (100°C - 0°C) = 100 K

ΔHtemp = 10.0 g x 0.138 J/g*K x 100 K = 138 J = 0.138 kJ

Now we can add up all the terms to get the total enthalpy required:

ΔH = 3.94 kJ + 0.138 kJ = 4.08 kJ

Therefore, the closest answer is A) 6.8 kJ.

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