Question

1. Butane (C4H10) has the following properties: Tm = -140 oC, Tb = 1oC, ΔHfus = 4.66 kJ/mol, ΔHvap = 22.44 kJ/mol, cliquid = 132.42 J/molK, cgas = 98.49 J/molK, csolid = 85.74 J/molK

a. How much energy is required to vaporize 6.05 moles of butane at a constant temperature?

b. How much energy is released when 4.76 g of butane freezes?

c. How much energy is required to convert 10.5 g of solid butane at -145 oC to a gas at 5oC?

d. if 36.0 kJ of energy is removed from a 21.4 g sample of butane at 10^oC, what temperature will it end up at? Hint – it will be a liquid at the end and be below 1^oC.

Answer 1

1

a. energy required to vaporize = n*DHvap

                                = 6.05*22.44

                                 = 135.76 kj

b. energy released = n*dhFUS

                     = (4.76/58)*(4.66)

                     = 0.3824 Kj

C. energy is required (q) = nsolid*Csolid*DT1+n*DHfus+nliquid*Cliquid*DT2+n*DHvap+ngas*Cgas*DT3

                           = (10.5/58)*85.74*(-140-(-145))+(10.5/58)*4.66*10^3+(10.5/58)*132.42*(1-(-140))+(10.5/58)*22.44*10^3+(10.5/58)*98.49*(5-1)

                           = 8435.1 joule

                           = 8.435 kj

d. pls chek it one again , the given data

energy is required (q) = nliquid*Cliquid*DT2+n*DHvap+ngas*Cgas*DT1
        
         36*10^3 = (21.4/58)*132.42*(1-x)+(21.4/58)*22.44*10^3+(21.4/58)*98.49*(10-1)