The molar enthalpy of vaporization of boron tribromide is 30.5 kJ/mol, and its normal boiling point is 91°C. What is the vapor pressure of BBr₃ at 20.°C? (R = 8.314 J/K • mol) A) 11.5 torr B) 659 torr C) 255 torr D) 66.2 torr E) 167 torr

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To find the vapor pressure of boron tribromide (BBr3) at 20°C, we can use the Clausius-Clapeyron equation, which relates the change in vapor pressure with temperature to the enthalpy of vaporization. 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:- $\Delta H_{vap} = 30.5 \, \text{kJ/mol} = 30500 \, \text{J/mol}$ (conversion from kJ to J)- $R = 8.314 \, \text{J/K} \cdot \text{mol}$- $T_1 = 91°C = 364 \, K$ (boiling point, converted from °C to K by adding 273)- $T_2 = 20°C = 293 \, K$ (converted from °C to K by adding 273)- $P_1 = 760 \, \text{torr}$ (normal boiling point pressure)Rearranging the equation to solve for $P_2$, and substituting the given values:$$ \ln\left(\frac{P_2}{760}\right) = -\frac{30500}{8.314} \left(\frac{1}{293} - \frac{1}{364}\right) $$Solving this equation gives the value of $P_2$ in torr, which is approximately 66.2 torr.

The Molar Enthalpy of Vaporization of Boron Tribromide Is 30