Suppose that 1.048 g of benzene is combusted in a bomb calorimeter which contains 945 g of

water. The temperature of the water increases from 23.640 OC to 32.681 oC. The heat capacity of the
empty calorimeter is 891 J/ OC .
C6H6 (l) + 15/2 O2 (g) > 3 H2O (l) + 6 CO2 (g)
(a) Calculate DE for the combustion of benzene in kJ/mol .
(b) Calculate DH for the combustion of benzene in kJ/mol. Assume that the volumes of all liquids are
negligible compared to the volumes of the gases and that the temperature is 298 K.

To solve this problem, we'll use the equation q = mCΔT, where q is the heat absorbed or released in Joules, m is the mass in grams, C is the specific heat capacity in J/g°C, and ΔT is the change in temperature in °C.

First, let's calculate the heat absorbed by the water:

q_water = m_water * C_water * ΔT_water

Given:
m_water = 945 g
C_water = 4.18 J/g°C (specific heat capacity of water is approximately 4.18 J/g°C)
ΔT_water = (32.681°C - 23.640°C)

q_water = 945 g * 4.18 J/g°C * (32.681°C - 23.640°C)
q_water = 93858.111 J

Next, let's calculate the heat absorbed by the empty calorimeter:

q_calorimeter = C_calorimeter * ΔT_calorimeter

Given:
C_calorimeter = 891 J/°C
ΔT_calorimeter = (32.681°C - 23.640°C)

q_calorimeter = 891 J/°C * (32.681°C - 23.640°C)
q_calorimeter = 8019.111 J

Now, let's calculate the total heat released by the combustion of benzene:

q_total = q_water + q_calorimeter

q_total = 93858.111 J + 8019.111 J
q_total = 101877.222 J

To convert this to kJ/mol, we need to calculate the moles of benzene.

Given:
m_benzene = 1.048 g
Molar mass of benzene (C6H6) = 78.11 g/mol

moles_benzene = m_benzene / Molar mass of benzene
moles_benzene = 1.048 g / 78.11 g/mol
moles_benzene = 0.0134 mol

Now, let's calculate ΔE for the combustion of benzene:

ΔE = q_total / moles_benzene
ΔE = 101877.222 J / 0.0134 mol
ΔE = 7603.753 kJ/mol (rounding to four decimal places)

Finally, to calculate ΔH for the combustion of benzene at 298 K, we can use the equation ΔE = ΔH - ΔnRT.

Given:
R = 8.314 J/mol·K
T = 298 K
Δn (change in moles of gas) = (moles of products - moles of reactants)

For the balanced equation: C6H6 (l) + 15/2 O2 (g) -> 3 H2O (l) + 6 CO2 (g)
Δn = (3+6) - (1+15/2) = 5/2

ΔE = ΔH - ΔnRT
7603.753 kJ/mol = ΔH - (5/2) * 8.314 J/mol·K * 298 K

Now, solve for ΔH:

ΔH = 7603.753 kJ/mol + (5/2) * 8.314 J/mol·K * 298 K
ΔH = 7603.753 kJ/mol + 6162.995 kJ/mol
ΔH = 13766.748 kJ/mol (rounding to four decimal places)

Therefore, the answers to the given questions are:
(a) ΔE for the combustion of benzene is 7603.753 kJ/mol.
(b) ΔH for the combustion of benzene is 13766.748 kJ/mol.

To calculate the enthalpy change (ΔH) for the combustion of benzene, you can use the equation:

ΔH = q + nCΔT

where ΔH is the enthalpy change, q is the heat absorbed or released in the reaction, n is the number of moles of substance being reacted, C is the heat capacity of the calorimeter, and ΔT is the change in temperature.

(a) To calculate ΔE for the combustion of benzene in kJ/mol:

1. Convert the mass of benzene to moles. The molar mass of benzene (C6H6) is 12.01 g/mol (carbon) + 1.01 g/mol (hydrogen) = 78.11 g/mol. Therefore, 1.048 g of benzene is equal to 1.048 g / 78.11 g/mol = 0.01342 mol.

2. Calculate the heat absorbed or released (q) by the water. The heat absorbed or released by the water can be calculated using the formula:

q = m × C × ΔT

where m is the mass of the water, C is the heat capacity of the water (assumed to be 4.18 J/g·°C), and ΔT is the change in temperature. Rearranging the equation, we get:

q = m × C × ΔT = 945 g × 4.18 J/g·°C × (32.681 °C - 23.640 °C) = 37923.23 J

3. Convert the heat absorbed or released to kJ by dividing by 1000:

q = 37923.23 J / 1000 = 37.92323 kJ

4. Calculate ΔE for the combustion of benzene by rearranging the equation and substituting the values:

ΔE = q / n = 37.92323 kJ / 0.01342 mol ≈ 2827 kJ/mol

Therefore, the ΔE for the combustion of benzene is approximately 2827 kJ/mol.

(b) To calculate ΔH for the combustion of benzene in kJ/mol:

To calculate ΔH, we need to account for the heat absorbed or released by the empty calorimeter. Assume that the heat capacity of the empty calorimeter is C_cal and the change in temperature is ΔT_cal. Therefore:

ΔH = ΔE + nC_calΔT_cal

Given that the heat capacity of the empty calorimeter (C_cal) is 891 J/°C and the change in temperature (ΔT_cal) is the same as the change in temperature of the water (32.681 °C - 23.640 °C), we can substitute the values into the equation:

ΔH = 2827 kJ/mol + 0.01342 mol × 891 J/°C × (32.681 °C - 23.640 °C)

Calculating this expression will give you the final value of ΔH for the combustion of benzene in kJ/mol.

At constant volume you calculate dE.

dE = [(mass H2O x specific heat H2O x (Tfinal-Tinitial)] + Ccal*(Tfinal-Tinitial)
Then dE x (g C6H6/molar mass C6H6) = J/mol. Convert to kJ/mol

For dH, use dE = dH + w and w can be obtained from pdV = delta n(RT)
You will need to assign the proper sign to work.