How much heat is required to convert 30.0 g of ethanol at 25 ° C to the vapor ph

Question

How much heat is required to convert 30.0 g of ethanol at 25 ° C to the vapor phase at 78 °C? Express your answer using two significant figures Constants | Periodic Table Ethanol (C2 Hs OH) melts at -114°C and boils at 78 °C The enthalpy of fusion of ethanol is 5.02 kJ/mol, and its enthalpy of vaporization is 38.56 kJ/mol. The specific heat of solid and liquid ethanol are 0.97 J/g.HK are 2.3 J/g K respectively. Q- k.J Submit Part B How much heat is required to convert 30.0 g of ethanol at -143 °C to the vapor phase at 78 °C? Express your answer using two significant figures. kJ

Explanation / Answer

PART A:

In contains two step process:

Step 1: Heat required for Liquid ethanol at 25 oC to liquid ethanol 78 oC = q1

Step 2: Heat required for Liquid ethanol at 78 oC to vapor ethanol 78 oC = q2

q1 = m C dT
     = (30 g) x (2.3 J g-1 K-1) x (78 oC – 25 oC)
     = (30 g) x (2.3 J g-1 K-1) x (53 oC)
     = (30 g) x (2.3 J g-1 K-1) x (53 K)     change in oC temperature = K temperature
     = 3657 J

q2 = m dHvap
    = (30 g / 46 g/mol) x (38.56 kJ/mol)
    = (0.65 mol) x (38560 J/mol)
   = 25064 J

So, total heat = 3657 J + 25064 J
                      = 28721 J
                      = 28.72 kJ
= 29 kJ (upto 2 sig. dig)

PART B:

In contains two step process:

Step 1: Heat required for Solid ethanol at - 143 oC to Solid ethanol - 114 oC = q1

Step 2: Heat required for Solid ethanol at - 114 oC to liquid ethanol - 114 oC = q2

Step 3: Heat required for liquid ethanol at - 114 oC to liquid ethanol 78 oC = q3

Step 4: Heat required for Liquid ethanol at 78 oC to vapor ethanol 78 oC = q4

q1 = m C dT
     = (30 g) x (0.97 J g-1 K-1) x (- 114 oC – (-143) oC)
     = (30 g) x (0.97 J g-1 K-1) x (29 oC)
     = (30 g) x (0.97 J g-1 K-1) x (29 K)     change in oC temperature = K temperature
     = 843.9 J

q2 = m dHfus
    = (30 g / 46 g/mol) x (5.02 kJ/mol)
    = (0.65 mol) x (5020 J/mol)
   = 3263 J

q3 = m C dT
     = (30 g) x (2.3 J g-1 K-1) x (78 oC – (-114) oC)
     = (30 g) x (2.3 J g-1 K-1) x (192 oC)
     = (30 g) x (2.3 J g-1 K-1) x (192 K)     change in oC temperature = K temperature
     = 13248 J

q4 = m dHvap
    = (30 g / 46 g/mol) x (38.56 kJ/mol)
    = (0.65 mol) x (38560 J/mol)
   = 25064 J

So, total heat = 843.9 J + 3263 J + 13248 J + 25064 J
                      = 42418.9 J
                      = 42.4189 kJ
                      = 42 kJ      (upto 2 sig. dig.)

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