Calculating the enthalpy and internal energy change for combustion

Question

The combustion of 1.899 g of phenol, C6H5OH(s), in a bomb calorimeter with a heat capacity of 4.80 kJ/°C results in an increase in the temperature of the calorimeter and its contents from 22.70 °C to 35.5 °C.

a)Calculate the enthalpy of combustion for phenol in kilojoules per mole.

b) What is the internal energy change for the combustion of 1.899 g of phenol in the bomb calorimeter?



I understand that for part a it is -3044.824 kj/mol but I don't understand part b.

Answer 1

1.899 g x 1 mol/94.11 g = 0.0218 moles phenol

-3044.824 kJ/mol x 0.0218 moles = 61.44 kJ = qΔ

ΔU = q + w = q -PΔV

At contant P, ΔU = q = 61.44 kJ (negative)

Answer 2

The general equation for internal energy change (ΔU) in a closed system is:

ΔU = Q + W_PV + W_Isochoric

W_PV is work related to pressure and volume changes. In this case neither is present (if the combustion were used to drive a movable piston, there would be some work involved).

W_Isochoric is work related to non-pressure/volume changes such as forcing electrons through a wire. Again this is not present here.

Q is the heat flow which we are dealing with in this combustion reaction.

Assuming your math is solid on part a), you could find the heat released by combustion of 1.899g of phenol and since combustion is exothermic, Q would be negative as heat is lost to the surroundings. So for this system and these conditions ΔU = Q.