Question

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 = 8.314j/ k • mol)?

Answer 1

To answer this question, we will use the following equation:
ln(P2/P1) = (∆Hvap/R)*((1/T1) - (1/T2)) 

Now we examine the givens of the problem and transform to standard units if required:
∆Hvap = 30.5 kJ/mol 
R is a constant = 8.314 x 10^-3 kJ K^-1 mol^-1 
T1 = 91 celcius = 91 + 273= 364 Kelvin
T2 = 20 celcius = 20 + 273 = 293 k3lvin
P1 is the standard atmospheric pressure = 760 mmHg
P2 is the value to be calculated

Substitute with these values in the equation:
ln(P2/760) = (30.5 / 8.314 x 10^-3) x ((1 / 364) - (1 / 293)) 
ln(P2/760) = - 2.4662 (Take the exponential both sides to eliminate the ln)
P2 / 760 = e^(-2.4462) = 0.0866
P2 = 0.0866 x 760 = 65.816 mmHg