Question 3 a) (6 points) In a constant-volume calorimetric bomb, we combust 1.77

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Question 3 a) (6 points) In a constant-volume calorimetric bomb, we combust 1.77g of C6H12(1). (Combustion is a reaction where a substance reacts with O2(g) in order to produce CO2 and H20(l)). The calorimetric bomb has a calorific capacity of 2.50 kJ K1 and contains 3.125 Kg of water. The specific heat capacity of water is 4.184 JK1g1. The temperature of the calorimeter and of the water raises from 21.25°C to 26.40?. Soon after, we combust 1.000 mol of the same substance, C6H12(1l) at a constant pressure of 1.00 atm at 25.0°C. Calculate the values of Q, W, ??, ?? and the variation of entropy of the surroundings during this combustion of 1.00 moles under constant pressure. b) (2 points) The vaporization enthalpy of water is 40.65 kJ/mol. What would be the variation of entropy of 1.000 mol when it boils under a constant pressure of 1.00atm?

Explanation / Answer

The combustion reaction is represented as C6H12(l)+ 9O2(g) ---->6CO2 (g)+ 6H2O (l)

Since the reaction is combustion and is exothermic. So enthalpy change, deltaH is –ve.

For constant pressure process, deltaH=Q, heat generated

The heat generated is given to water and calorimeter

Heat generated, Q= -( mass of water* specific heat of water* change in temperature+ heat capacity of calorimeter* change in temperature) (1)

Since specific heat of water = 4.184 J/gm.deg.c. expressing specific heat in Kj/kg.deg.c

Specific heat of water= 4.184*10-3 KJ/10-3 kg.deg.c= 4.184 Kj/kg.deg.c

Mass of water =3.125 kg, substituting these values in Eq.1

Gives= -{3.125Kg* 4.184 Kj/kg.deg.c*(26.40-21.25)+2.5Kj/K*(26.4-21.25)}=-80.21 Kj

This is the heat liberated or enthalpy change when 1.77 gm of C6H12 is burnt. Moles of C6H12=

Mass/molar mass, molar mass of C6H12=6* atomic weight of C+12* atomic weight of H

Atomic weights ( g/mole): C=12 and H= 1. Hence molar mass of C6H12= 6*12+12= 84 g/mole

mass/molar mass= 1.77/84=0.021071

hence enthalpy change/ mole= Q = -80.21 KJ/0.021071= -3806 Kj/mole

since ,deltaH= deltaU+delta(PV)

deltaH= deltaU+deltan*R*T, deltan= change in no of moles of gaseous species= 6-9= -3

deltaU= deltaH-deltan*T*R = -3806+3*8.314J/mole.*10-3 Kj/J*298 =-3799 Kj/mole

from 1st law of thermodynamics, deltaU= Q+W

W= deltaU-Q=-3799+3806=7 Kj/mole

Entropy change of system = Q/T= -3806/298 Kj/K=-12.77 Kj/K

Since the system is adiabatic, entropy of surroundings= -entropy of system=12.77 Kj/K

2. At 1 atm, the boiling point of water= 298K

Since during vaporization , heat is absorbed, deltaH is +ve.

So entropy of vaporization = enthalpy change/ boiling point =40.65 Kj/mole/298 =0.14 Kj/mole

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