A sample of biphenyl (C_6H_5)_2 weighing 0.526 g was ignited in a bomb calorimet

Question

A sample of biphenyl (C_6H_5)_2 weighing 0.526 g was ignited in a bomb calorimeter initially at 25 degree C, producing a temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid C_6H_5COOH weighing 0.825 g was ignited under identical conditions and produced a temperature rise of 1.94 K. For benzoic acid, the heat of combustion at constant pressure is known to be 3226 k) mol6-1 (that is, Delta U = -3226 kJ mol^-1.) Use this information to determine the standard enthalpy of combustion of biphenyl.

Explanation / Answer

m = 0.526 g of biphenyl

mol = mass/M W= 0.526/154.21 = 0.0034109 mol of biphenyl

dT = +1.91 K

benzoic acid = 0.825 g

mol of BA = mass/MW = 0.825/122.123 = 0.006755 mol

dT = 1.94

note that

HRXDN BA = 3226 kJ/mol

use this to determine the combustion of biphneyl

then

get Ccal

Qcal = HRxn*n = 0.006755 *3226 = 21.79163kJ

Ccal = Qcal/dT = 21.79163 / 1.94 = 11.232 kJ/°C

now

for biphenyl

Qcal = Ccal * d T= 11.232 * 1.91 = 21.45312 kJ

now

HRxn = -Qcal/n = 21.45312 / 0.0034109

HRxn = 6289.577 kJ/mol biphenyl

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.