## Chemistry and Chemical Reactivity (9th Edition)

From the Clausius-Clapeyron equation we obtain: $ln\dfrac{P_2}{P_1}=-\dfrac{\Delta H_v}{R}×\left(\dfrac 1{T_2}-\dfrac 1{T_1}\right)$ $ln(29.70/14.70)=-44.0\ kJ/mol/0.008314\ kJ/mol.K×(1/T_2-1/373)$ Solving for the temperature gives: $T=392\ K=119.5°C$