Question

A 1.00 g sample of n-hexane (C6H14) undergoes complete combustion with excess O2 in a bomb calorimeter. The temperature of the 1502 g of water surrounding the bomb rises from 22.64°C to 29.30°C. The heat capacity of the hardware component of the calorimeter (everything that is not water) is 4042 J/°C. What is ΔU for the combustion of n-C6H14? One mole of n-C6H14 is 86.1 g. The specific heat of water is 4.184 J/g·°C.

Answer 1

Answer:

-5,921x10⁶J/mol

Explanation:

Internal energy change (ΔU) for the reaction of combustion in the bomb calorimeter is:

ΔU = q calorimeter + q solution

Where:

q calorimeter is Ccal×ΔT (Ccal=4042J/°C) and (ΔT is 29,30°C-22,64°C=6,66°C)

q solution is c×m×ΔT (c= 4.184 J/g°C), (m=1502g H₂O), (ΔT is 29,30°C-22,64°C=6,66°C)

Replacing:

ΔU = 26920J + 41854J = 68774 J

This energy is per g of n-hexane, now, per mole of n-hexane:

$\frac{68774J}{1gHexane} *\frac{86,1g}{1mol}$= -5,921x10⁶J/mol

-negative because the energy is produced-

I hope it helps!

Answer 2

Answer:

The correct answer is 5921.4 KJ/mol

Explanation:

The combustion of n-hexane in a bomb calorimeter is an adiabatic process, which means that the heat released by the reaction (q reaction) is absorbed by the calorimeter (q absorbed). The heat is absorbed in part by the hardware component of the calorimeter (Ccal= 4042 J/ºc) and in part by the surrounding water (Cw= 4.184 J/ºC). In a process at constant volume, the heat is equal to the variation of internal energy (ΔU). Thus, we have:

ΔU= - q reaction = q absorbed = qcal + qw

We calculate first the heat absorbed by the calorimeter (qcal):

qcal= Ccal x ΔT

qcal= 4042 J/ºC x (29.30ºC - 22.64ºC)

qcal= 26919.72 J

Secondly, we calculate the heat absorbed by the water (qw)

qw= m x Cw x ΔT

qw= 1502 g x 4.184 J/g.ºC x (29.30ºC - 22.64ºC)

qw= 41853.9 J

q absorbed = qcal + qw = 26919.72 + 41853.9 J = 68773.61 J

This amount of heat is released by 1 g of n-hexane. To calculate ΔU we use the molecular weight of n-hexane (86.1 g/mol):

ΔU = q absorbed = 68773.61 J/ 1 g x 86.1 g/ mol x 1 KJ/ 10³ J = 5921.4 KJ/mol