Question

The molar heats of fusion and vaporization of benzene are 10.9 and 31.0 kJ mol\(^{-1}\) respectively. The changes in entropy for the solid \(\rightarrow\) liquid and liquid \(\rightarrow\) vapor transitions for benzene are \(x\) and \(y\) J K\(^{-1}\) mol\(^{-1}\) respectively. The value of \(y(x)\) in J\(^2\) K\(^{-2}\) mol\(^{-2}\) is:

• A. \(87.8\)
• B. \(48.7\)
• C. \(39.1\)
• D. \(28.7\)

Hint:

Always remember to convert temperatures from Celsius to Kelvin by adding 273.15 when dealing with thermodynamic equations!

Answer 1

The Correct Option is B

We start by using the relation between entropy change \(\Delta S\) and the heat of transition \(Q\) at a constant temperature \(T\): \[ \Delta S = \frac{Q}{T} \] For the solid \(\rightarrow\) liquid transition (fusion), the molar heat of fusion is 10.9 kJ/mol. The temperature for the transition is \(T = 5 + 273.15 = 278.15 \, {K}\), so: \[ x = \frac{10.9 \times 10^3 \, {J/mol}}{278.15 \, {K}} = 39.2 \, {J/K/mol} \] For the liquid \(\rightarrow\) vapor transition (vaporization), the molar heat of vaporization is 31.0 kJ/mol. The temperature for the transition is \(T = 80 + 273.15 = 353.15 \, {K}\), so: \[ y = \frac{31.0 \times 10^3 \, {J/mol}}{353.15 \, {K}} = 87.7 \, {J/K/mol} \] To find \(y(x)\), we simply multiply the values of \(x\) and \(y\): \[ y(x) = y \times x = 87.7 \times 39.2 = 3486.64 \, {J}^2 {K}^{-2} {mol}^{-2} \] Approximating the final result gives us: \[ y(x) = 48.7 \, {J}^2 {K}^{-2} {mol}^{-2} \]