Could you please explain simply step by step so I can understand it ? 4. Calcula

Question

Could you please explain simply step by step so I can understand it ? 4. Calculate the amount of energy (in kJ) to be required for a 132-g sample of benzene (CaHo 'M 78.1 15 g/mol) at 680°C to rise to 1520°C? The normal freezing point for benzene is 5.85 °C and boiling point is 79.85 °C. Physical data associated with benzene is provided below, but all of the data may not need to be used. (8 pts.) Cs, solid= 1.51 J/g°C AHfus = 9.80 kJ/mol Cs, liquid = 1.73 J/g·°C ar,a 33.9 kJ/mol Cs. gas 1.06 J/g. °C Sara 132 17 : 57, 241 K3 (y

Explanation / Answer

Ti = 68.0 oC

Tf = 152.0 oC

here

Cl = 1.73 J/g.oC

Heat required to convert liquid from 68.0 oC to 79.85 oC

Q1 = m*Cl*(Tf-Ti)

= 132 g * 1.73 J/g.oC *(79.85-68) oC

= 2706.066 J

Lv = 33.9KJ/mol =

33900J/mol

Lets convert mass to mol

Molar mass of C6H6 = 78.115 g/mol

number of mol

n= mass/molar mass

= 132.0/78.115

= 1.69 mol

Heat required to convert liquid to gas at 79.85 oC

Q2 = n*Hvap

= 1.69 mol *33900 J/mol

= 57289.9063 J

Cg = 1.06 J/g.oC

Heat required to convert vapour from 79.85 oC to 152.0 oC

Q3 = m*Cg*(Tf-Ti)

= 132 g * 1.06 J/g.oC *(152-79.85) oC

= 10095.228 J

Total heat required = Q1 + Q2 + Q3

= 2706.066 J + 57289.9063 J + 10095.228 J

= 70091 J

= 70.1 KJ

Answer: 70.1 KJ

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