How much heat energy is required to convert 93.4 g of solid ethanol at − 114.5 ° c to gasesous ethanol at 149.8 ° c ? the molar heat of fusion of ethanol is 4.60 kj/mol , and its molar heat of vaporization is 38.56 kj/mol . ethanol has a normal melting point of − 114.5 ° c and a normal boiling point of 78.4 ° c . the specific heat capacity of liquid ethanol is 2.45 j / g ⋅ ° c , and that of gaseous ethanol is 1.43 j / g ⋅ ° c .

Q' = 140.859 kJ

Explanation:

Given that, 93.4 g of solid ethanol at − 114.5 °C is converted to gasesous ethanol at 149.8 ° C.

The molar heat of fusion of ethanol is, ΔH(f) = 4.60 kJ/mol , and its molar heat of vaporization is ΔH(v) = 38.56 kJ/mol .

And also Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C .

The specific heat capacity of liquid ethanol is S(l) = 2.45 J / g ⋅°C , and that of gaseous ethanol is S(g) = 1.43 J / g ⋅°C .

Lets solve this step wise ;

Given 93.4 g of ethanol is taken, but 1 mole of ethanol consists of 46.06 g

⇒ moles of ethanol given = = 2.02 moles

step 1: solid ethanol to liquid ethanol at melting point of − 114.5 ° C

⇒ 1 mole requires ΔH(f) = 4.60 kJ/mol of heat

⇒ heat required = 4.60 × 2.02 = 9.292 kJ.

step 2: liquid ethanol at -114 °C to liquid ethanol at 78.4 °C

Q = m×S×ΔT ; Q = heat required

                       m = mass of the substance

                       S = specific heat of the substance

                       ΔT = change in temperature

Here S = S(l);

⇒ Q = 93.4×2.45×(78.4-(-114.5))

       = 44.141 kJ

step 3: liquid ethanol at 78.4°C to gaseous ethanol at 78.4°C

1 mol of liquid ethanol requires ΔH(v) = 38.56 kJ/mol of heat

⇒ required heat = 38.56×2.02 = 77.89 kJ

step 4: gaseous ethanol at 78.4 °C to gaseous ethanol at 149.8 °C

Q = m×S×ΔT

Here, S = S(g)

Q = 93.4×1.43×(149.8-78.4)

   = 9.536 kJ

⇒ Total heat required = 9.292 + 44.141 + 77.89 + 9.536

                                     = 140.859 kJ

⇒ Q' = 140.859 kJ

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