Question 5.4.4: Use constant-volume calorimetry to determine energy change. ......

You are asked to calculate the energy change (in kJ/mol sucrose) for the combustion of sucrose.

You are given the mass of sucrose and bomb calorimeter data (mass of the water, temperature change of the water, and calorimeter heat capacity).
First, calculate the energy absorbed by the water bath and the bomb.
Water bath:

c_{water} = 4.184 J/g · ºC

m_{water} = 748 g

∆T = T_{final} − T_{initial} = 22.06 ºC − 20.00 ºC = 2.06 ºC

q_{water} = m × c_{water} × ∆T = (748 g)(4.184 J/g · ºC)(206 ºC) = 6450 J

Bomb:

c_{bomb} = 420. J/ºC

∆T = T_{final} − T_{initial} = 22.06 ºC − 20.00 ºC = 2.06 ºC

q_{bomb} = c_{bomb} × ∆T = (420. J/g · ºC)(2.06 ºC) = 865 J

Next, calculate the energy released by the combustion reaction

0 = q_{reaction} + q_{bomb} + q_{water}

q_{reaction} = −(q_{bomb} + q_{water}) = −(6450 J + 865 J) = −7320 J

Finally, calculate the amount (in mol) of sucrose burned in the combustion reaction and the energy change for the reaction.

0.444 g C_{12}H_{22}O_{11} × \frac{1\text{ mol C}_{12}\text{H}_{22}\text{O}_{11}}{342.3\text{ g}} = 0.00130 mol C_{12}H_{22}O_{11}

ΔE = \frac{q_{reaction}}{\text{ mol C}_{12}\text{H}_{22}\text{O}_{11}} = \frac{-7320\text{ J}}{0.00130\text{ mol}} \times \frac{1\text{ KJ}}{10^{3}\text{ J}} =  -5640 kJ/mol

Because the difference between ∆E and ∆H is quite small, this energy change can be taken as the enthalpy of combustion (∆H_{comb}) for sucrose.

Is your answer reasonable? This is an exothermic reaction; temperature increased, and q_{reaction} and ∆E are negative. In addition, the combustion of a hydrocarbon typically produces a large amount of energy per mole of compound.