An equimolar liquid mixture of benzene and toluene at 10C is fed continuously to
ID: 1069875 • Letter: A
Question
An equimolar liquid mixture of benzene and toluene at 10C is fed continuously to a vessel in which mixture is heated to 50C. The liquid product is 40 mole%B and the vapor product is 68.4% B. How much heat must be transferred to the mixture per mol of feed? An equimolar liquid mixture of benzene and toluene at 10C is fed continuously to a vessel in which mixture is heated to 50C. The liquid product is 40 mole%B and the vapor product is 68.4% B. How much heat must be transferred to the mixture per mol of feed?Explanation / Answer
Basis: 1 mole of feed
Total moles of benzene=0.5 moles
Total moles of toluene=0.5 moles
After heating the feed , let the total moles present in liquid phase is N and in gas pahse is 1-N ( since total moles are 1)
Mole balance on Toluene(B)
in=out
0.5= 0.4*N+(1-N)*0.684
Solving for N gives .6478
i.e total moles in liquid phase are 0.6478 and that of in vapor phase are 0.3522
Moles of benzene in liquid phase = 0.6*0.6478=0.38873
Moles of Toluene in liquid phase = 0.4*0.6478=0.25912
Moles of benzene in vapor phase = 0.316*0.3522=0.1112
Moles of Toluene in vapor phase = 0.684*0.3522=0.2409
latent heat of benzene at 50 c= 32.385 kj/mole
latent heat of toluene at 50 C=41.94kj/mole
specific heats of benzene and toluene are 134.8 j/mole/k and 155.96 j/mol/k
Heat supplied = heat supplied to liquid benzene to rise the temp from 10 to 50 oC + latent heat of vaporization of benzene at 50 oc+ heat supplied to liquid Toluene to rise the temp from 10 to 50 oC + latent heat of vaporization of Toluene at 50 oc
= 0.5*134.8*(50-10) + (32.385*1000)*0.1112 +0.5*155.96*(50-10)+(41.94*1000)*0.2409=101042876 Joules
{Note : latent heats of pure substances are readily available in the literature at normal boiling points. To calculate latent heats at 50 C , make use of watson equation.}