Bromine has pressure at vapourization = 400mm at 41.0 degrees Celsius and a normal boiling point of 331.9K. What is the heat of vaporization, delta Hvap of bromine in kJ/mol?

Asked on by sere-mill

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llltkl | College Teacher | (Level 3) Valedictorian

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From the Clausius-Clapeyron relation, as applied to the vaporization:
`ln(P_1/P_2) = ((DeltaH_(vap))/R)*(1/T_2-1/T_1)`
`P_1=400 mm` , `P_2=760 mm` , `T_1` =(273+41), i.e. `314` K, and `T_2=331.9` K

Plugging in the values, we get:
`DeltaH_(vap)= R*ln(P_1/P_2)/(1/T_2-1/T_1)`
`= 8.3145*ln(400/760)/(1/331.9-1/314)`
`= 31.07 (kJ)/(mol)`


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