Given the data below,determine the molar enthalpy of vaporization of COCl₂. COCl₂,P₁ = 40 mmHg,t₁ = -50.3 °C,P₂ = 100 mmHg,t₂ = -35.6 °C A)0.518 kJ/mol B)27.4 kJ/mol C)4.32 kJ/mol D)0.928 kJ/mol E)0.112 kJ/mol

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The Clausius-Clapeyron equation can be used to calculate the molar enthalpy of vaporization ($\Delta H_{vap}$) using the given data. The equation in its linear 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:- $P_1 = 40$ mmHg, $T_1 = -50.3$ °C $= 222.85$ K (since $T(K) = T(°C) + 273.15$)- $P_2 = 100$ mmHg, $T_2 = -35.6$ °C $= 237.55$ KFirst, convert pressures to the same units if necessary (in this case, both are in mmHg, so no conversion is needed). Then, solve for $\Delta H_{vap}$:$$\ln\left(\frac{100}{40}\right) = -\frac{\Delta H_{vap}}{8.314}\left(\frac{1}{237.55} - \frac{1}{222.85}\right)$$Solving for $\Delta H_{vap}$ gives a value close to 27.4 kJ/mol, indicating that the correct choice is B.

Given the Data Below,determine the Molar Enthalpy of Vaporization of COCl2

Given the Data Below,determine the Molar Enthalpy of Vaporization of COCl₂