The normal boiling point of \(\mathrm{Br}_{2}(l)\) is \(58.8^{\circ} \mathrm{C},\) and its molar enthalpy of vaporization is \(\Delta H_{\mathrm{vap}}=29.6 \mathrm{kJ} / \mathrm{mol}\) (a) When \(\mathrm{Br}_{2}(l)\) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when 1.00 mol of \(\mathrm{Br}_{2}(l)\) is vaporized at \(58.8^{\circ} \mathrm{C}\) .

When a substance undergoes a phase change, such as boiling, the particles become more disordered and more dispersed. In the case of liquid bromine boiling and becoming gaseous bromine, the increased freedom of movement and spacing between the particles contribute to a higher entropy, which is a measure of disorder in a system. Thus, during the boiling of Br2(l) at its normal boiling point, the entropy increases.

Next, we calculate the entropy change (∆S) when 1.00 mol of Br2(l) is vaporized at its boiling point. We use the equation: \(\Delta S = \frac{\Delta H_{vap}}{T}\) We are given the molar enthalpy of vaporization \(\Delta H_{vap} = 29.6 \, kJ/mol\) and the boiling point temperature \(T = 58.8^{\circ} C\). First, we need to convert the temperature from Celsius to Kelvin: \(T_{(K)} = T_{(C)} + 273.15\) \(T_{(K)} = 58.8 + 273.15 = 331.95\,K\) Now, we apply the equation: \(\Delta S = \frac{29.6 \, kJ/mol}{331.95 \, K}\) Note that we need to convert kJ to J by multiplying with 1000: \(\Delta S = \frac{29.6 \times 1,000\, J/mol}{331.95 \, K} = 89.1\frac{J}{mol \cdot K}\) Therefore, the entropy change, ∆S, when 1.00 mol of Br2(l) is vaporized at its normal boiling point is 89.1 J/(mol·K).