Question

How much energy must be removed from a 125 g sample of benzene (C6H6) vapor at 425.0 K to make it a solid at 200.0 K? Use the following steps to get to your answer. You MUST show your work to receive any credit. The following physical data for benzene may be useful. The quiz will be due at the beginning of class on 2/17/16. ΔHvap = 33.9 kJ/mol ΔHfus = 9.8 kJ/mol Cliq = 1.73 J/g°C Cgas = c Csol = 1.51 J/g°C Tmelting = 279.0 K Tboiling = 353.0 K

7) What is the total amount of energy, in kJ, needed

Answer 1

Answer – We are given, mass of benzene = 125 g, ti = 425 K

tf = 200.0 K , ∆Hvap = 33.9 kJ/mol , ∆Hfus = 9.8 kJ/mol Cliq = 1.73 J/g°C Cgas =1.06 J/g°C   Csol = 1.51 J/g°C ,T melting = 279.0 K , T boiling = 353.0 K

moles of benzene = 125 g / 78.11 g.mol^-1

                             = 1.60 moles

Now first we need to calculate the energy form the 425 K to 353.0 K

We know,

q1 = m*C* Δt

= 125 g * 1.06 J/g°C *(353-425)

= -9540 J

Heat form the 353 K to 353 K

q2 = m*∆Hvap

    = 1.60 mol * 33900 J/mol

    = -54240 J

Heat form the 353 K to 279 K

q3 = m*C* Δt

= 125 g * 1.73 J/g°C *(279-353)

= -16002.5 J

Heat form the 279 K to 279 K

q4 = m*∆Hfus

    = 1.60 mol * 9800 J/mol

    = -15680 J

Heat form the 279 K to 200 K

q5 = m*C* Δt

= 125 g * 1.51 J/g°C *(200-279)

= -15010 J

Total heat removed for the benzen q = q1 + q2 + q3 +q4 +q 5

                                              = -9540 J -54240 J -16002.5 J – 15680 J – 15010 J

                                              = -1.1047*10⁵ J

                                              = -110.5 kJ

the total amount of energy, in kJ, needed = -110.5 kJ