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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 .

Respuesta :

Answer:

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 = [tex]\frac{93.4}{46.06}[/tex] = 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

snehashish65

The total heat energy required to convert 93.4 g of solid ethanol at − 114.5 ° C to gaseous ethanol at 149.8 ° C is 140.859 kJ.

Given data:,

93.4 g of solid ethanol at − 114.5 °C is converted to gaseous ethanol at 149.8 ° C.

The molar heat of fusion of ethanol is, ΔH(f) = 4.60 kJ/mol.

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 .

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

⇒ moles of ethanol given is

⇒93.04/46.06 =  2.02 moles

Heat required for conversion of solid ethanol to liquid ethanol at melting point of − 114.5 ° C

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

⇒ Q = 4.60 × 2.02 = 9.292 kJ.

Heat required to convert liquid ethanol at -114 °C to liquid ethanol at 78.4 °C

Q' = m×s×ΔT

Here,

m = mass of the substance

s = specific heat of the substance

ΔT = change in temperature

Solving as,

Q' = 93.4×2.45×(78.4-(-114.5))

Q' = 44.141 kJ

Heat required to convert 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

Q'' = 38.56×2.02 = 77.89 kJ

Heat required to convert gaseous ethanol at 78.4 °C to gaseous ethanol at 149.8 °C

Q''' = m×S×ΔT

 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

Thus, we can conclude that the total heat energy required to convert 93.4 g of solid ethanol at − 114.5 ° C to gaseous ethanol at 149.8 ° C is 140.859 kJ.

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