How much energy must be removed from a 125g sample of benzene(molar mass=88.11g/

Question

How much energy must be removed from a 125g sample of benzene(molar mass=88.11g/mol) at 425 k to liquify the sample and lowerthe temperature to 335.0k? The following physical data may beuseful. Hvap= 33.9 kj/mol Cliq= 1.73 J/g*C Cgas= 1.06 J/g*C Tboiling=353.0 K How much energy must be removed from a 125g sample of benzene(molar mass=88.11g/mol) at 425 k to liquify the sample and lowerthe temperature to 335.0k? The following physical data may beuseful. Hvap= 33.9 kj/mol Cliq= 1.73 J/g*C Cgas= 1.06 J/g*C Tboiling=353.0 K

Explanation / Answer

This is really just a big math problem. What you are dois first of all calculate the energy need into two parts. Thegas phase and the liquid phase. this is divided by the boilingpoint. So the gas phase is from 353-425K and the liquid phaseis from 335-353K convert these into celcius because you constant isin celcius. K=273+C gas is from 80 -152 Cand liquid is from 62-80. Then cacluate the J for eachphase. Gas 72 C * 1.06 J/g*C * 125 g= 9540 J Liquid 18 C * 1.73 J/g*C * 125 g=3892.5 J Then you add those plus the Hvap times the number ofmoles which is the energy it took to change phases. add them alltogether and you get 125 g /88.11 g/mol=1.4186 mol 1.4186 mol * 33.9 kJ/mol=48.1 kJ 9540+3892.5+48100= 61532.5J =61.5 kJ Which is your final answer

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