Question

C_gas = 1.3 J/g×K    H_vap = 38.56 kJ/mol    C_liq = 2.3 J/g×K      H_fus = 5.02 kJ/mol    C_solid = 0.97 J/g×K

     Condensation Temp = 78.0^oC               Freezing Pt = -114.0^oC

Calculate the amount of heat required toconvert 92.6 mL of ethanol, C₂H₆O, from 110.0^oC to -98.0^oC.

Answer 1

Assuming the denity of ethanol is constant at around 0.79 g/ml,
mass of ethanol present in 92.6 ml = ( 92.6 ml) x ( 0.79 g/ml) = 73.15 g (It would remain constant during the tranformation process)

mole of ethanol (n)= (73.15 / (46 g/mol)) = 1.59 or 1.6 mol

Processes taking place:

First At 110.0oC, ethanol is in gaseos state. Till 78oC , which is its Condensation Temperature, it will be in gaseous state.

= 78oC - 110.0oC = -32 oC = -32oC or -32 K    .

Note that difference in temperature would be same in both the units, oC or K

mass = 73.15 g , and Cgas = 1.3 J/g×K

q1 = m x C x = (73.15 g) x (1.3 J/g×K ) x ( -32 K) = -3043.04 J

q1 = -3.04 kJ

Second At 78 oC , it will condense to form liquid. During phase change, temperature remain constant, and energy released would be = n x Hvap = (1.6 ) x ( 38.56 kJ/mol ) = 61.32 kJ

q2 = -61.32 kJ

Note that Hvap is for the process: liquid gas,   q = Hvap

and here the reverse process that is condensation is taking place, gas liquid , q' = -Hvap ; or heat is released.

Third From 78oC to -98.0oC, the liquid is being cooled. Since the freezing temperature, when the liquid changes to solid is -114.0oC, is even lower tha -98.0 oC, ethanol will remain as a liquid.

= -98.0oC - 78oC = -176oC Or -176 K    .

mass = 73.15 g , and Cliquid = 2.3 J/g×K

q3 = m x C x = (73.15 g) x ( 2.3 J/g×K ) x ( -176 K ) = -29611.12 J

q3 = -29.6 kJ

Hence for overall process; q = q1 + q2 + q3

                                                       = (-3.04 kJ) + (-61.32 kJ) + (-29.6 kJ)

                                                = -93.96 kJ

or rounding upto 2 significant figures: q = -94 kJ

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