Did I correctly find the temperature of the vapor pressure using Clausius-Clapeyron equation?

Question

It has an enthalpy of vap of 41.098 kJ mol^-1. If the vapor pressure is 78.90 mmHg at 43.21 celsius, at what temp is vapor pressure equal to 666.6mmhg?

I used Clausius-Clapeyron formula:

ln(78.90mmhg/666.6mmhg)=41.098x10^-1kJ/0.008314kj/mol*K ((1/316.21K)-(1/t2)

and obtained: 133.6989 or 133.7K

is this correct?

Answer 1

Clausius Clapeyron Equation:

ln (P2/P1) = -Ea/R (1/T2 - 1/T1)

P1 = 78.90 mm Hg

P2 = 666.6 mm Hg

T1 = 316.2

T2 = ?

Ea = 41.098 kJ/mol

R = 0.008314 kJ/Kmol

ln(666.6/78.90) = - 41.098/0.008314 (1/T2 - 1/316.2)

2.13 = - 4943 (1/T2 - 0.00316)

2.13 = -4943/T2 + 15.63

13.50 = 4943/T2

T2 = 366K

(Please check the math, as I don't get the same answer as you)