The vapor pressure of nitric acid is 26.6 mmHg at 10.0 °C and 208 mmHg at 50.0 °C.Using this information,calculate the heat of vaporization (Δ_vapH)of nitric acid.(R = 8.314 J/K ⋅ mol)

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The Clausius-Clapeyron equation can be used to calculate the heat of vaporization ($\Delta H_{vap}$) from the vapor pressure at two different temperatures. The equation in its logarithmic form is:$$\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$Given:- $P_1 = 26.6$ mmHg at $T_1 = 10.0$ °C = 283.15 K- $P_2 = 208$ mmHg at $T_2 = 50.0$ °C = 323.15 K- $R = 8.314$ J/K ⋅ molFirst, convert pressures to the same units as R (if needed, but here we can cancel units since both pressures are in the same unit and we are interested in their ratio).$$\ln\left(\frac{208}{26.6}\right) = -\frac{\Delta H_{vap}}{8.314}\left(\frac{1}{323.15} - \frac{1}{283.15}\right)$$Solving for $\Delta H_{vap}$:$$\Delta H_{vap} = -\ln\left(\frac{208}{26.6}\right) \times \frac{8.314}{\left(\frac{1}{323.15} - \frac{1}{283.15}\right)}$$$$\Delta H_{vap} \approx 39.1 \, \text{kJ/mol}$$Therefore, the heat of vaporization of nitric acid is approximately 39.1 kJ/mol.

The Vapor Pressure of Nitric Acid Is 26