I need help with a chemistry question

Question

The compound trichlorosilane (SiCl₃H) has a vapor pressure of 

0.258 atm at −2°C. What is its normal boiling point? (The molar heat of vaporization of trichlorosilane is 

28.8 kJ/mol.)

Answer 1

_{We use the Clausius-Clapeyron equation,}

ln₍p_2​/p_{1​​) =} Δvap​H/R_{​ x (1/}T_1​​−1/T_2​)​ 

 

where,

p_{1 and} p_{2​ are the vapour pressures at temperatures} T_{1​ and} T₂

_Δvap​H _{= the enthalpy of vaporization of the liquid}

R _{= the Universal Gas Constant}

now at Normal Boiling point, P = 1atm

ln(1 atm / 0.258 atm) = 28800J/mol / 8.3145 J/molK x (1/271K - 1/T₂)

Solving for T₂, we get

T₂ = 303.13K

Converting Kelvin to Celsius,

0°C + 273= 273K

303.13K - 273K= 30.13°C

The normal boiling point of SiCl₃H is 30.13°C