How much energy must be removed from a 94.4 g sample of benzene (molar mass- 78.
ID: 536540 • Letter: H
Question
How much energy must be removed from a 94.4 g sample of benzene (molar mass- 78.11 g/mol) at 322.0 K to solidify the sample and lower the temperature to 205.0 K? The following physical data may be useful. (Delta H_vap = 33.9 kJ/mol, Delta H_fus = 9.8 kJ/mol, C_liq = 1.73 J/g degree C. C_gas = 106 J/g degree C. C_sol = 1.51 j/g degree C, T_melting = 279.0 K, T_boiling = 353.0 K) The fluorocarbon C_2 Cl_3 F_3 has a normal boiling point of 47.6 degree C. The specific heats of C_2 Cl_3 F_3 (1) and C_2 Cl_3 F_3 (g) are 0.91 J/gK and 0.67 J/gK. respectively. The heat of vaporization of the compound is 27.49 kJ/mol. How much heat (in kJ) is required to convert 50.0 g of the compound from the liquid at 5.0 degree C to the gas at 80.0 degree C?Explanation / Answer
11)
Ti = 322.0
Tf = 205.0
Cl = 1.73 J/goC = 1.73 J/gK
Heat released to convert liquid from 322.0 K to 279.0 K
Q1 = m*Cl*(Ti-Tf)
= 94.4 g * 1.73 J/gK *(322-279) K
= 7022.42 J
Lf = 9.8KJ/mol = 9800 J/mol
Lets convert mass to mol
Molar mass of C6H6 = 78.11 g/mol
number of mol
n= mass/molar mass
= 94.4/78.11
= 1.21 mol
Heat released to convert liquid to solid at 279.0 oC
Q2 = n*Lf
= 1.21 mol *9800 J/mol
= 11844.11 J
Cs = 1.51 J/goC = 1.51 J/gK
Heat released to convert solid from 279.0 to 205.0
Q3 = m*Cs*(Ti-Tf)
= 94.4 g * 1.51 J/gK *(279-205) K
= 10548.26 J
Total heat released = Q1 + Q2 + Q3
= 7022.42 J + 11844.11 J + 10548.26 J
= 29415 J
Answer: 9415 J or 9.415 KJ
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