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pychu [463]
3 years ago
9

How much heat (kj) is required to convert 3.12 moles of liquid benzene at 75.1°c to gaseous benzene at 115.1°c?

Chemistry
1 answer:
Tems11 [23]3 years ago
3 0
The total energy includes sensible heat to raise the temperature from 75.1°C to the boiling point. It also includes the latent heat to convert the liquid to gas. Then, it also includes sensible heat from he boiling point to 115.1°C. The equation is:

Energy = nCp,liquid(T,bp - T₁) + nΔH + nCp,gas(T₂ - T,bp)
where
n is the number of moles
T,bp is the boiling point of benzene at 80.1°C
Cp,liquid = 134.8 J/mol·°C
Cp,gas = 82.44 J/mol·°C
ΔH = 87.1 J/mol

Energy = (3.12 moles)(134.8 J/mol·°C)(80.1°C - 75.1°C) + (3.12 moles)(87.1 J/mol) + (3.12 moles)(82.44 J/mol·°C)(115.1°C - 80.1°C)
Energy = 11,377.08 J
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earnstyle [38]
Answer:
               Cp  =  0.093 J.g⁻¹.°C⁻¹
Solution:

The equation used for this problem is as follow,

                                                  Q  =  m Cp ΔT   ----- (1)

Where;
            Q  =  Heat  =  300 J

            m  =  mass  =  267 g

            Cp  =  Specific Heat Capacity  =  ??

            ΔT  =  Change in Temperature  =  12 °C

Solving eq. 1 for Cp,

                                 Cp  =  Q / m ΔT


Putting values,
                                 Cp  =  300 J / (267 g × 12 °C)

                                 Cp  =  0.093 J.g⁻¹.°C⁻¹
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