The heat of combustion of 2-propanol at 298.15 {\rm K}, determined in a bomb calorimeter, is -33.41{\rm{kJ/g}}.

(a)For the combustion of one mole of 2-propanol, determine \Delta U.

(b)For the combustion of one mole of 2-propanol, determine\Delta H.

To answer these questions, we need to understand the concepts of internal energy (\(\Delta U\)) and enthalpy change (\(\Delta H\)).

(a) \(\Delta U\) is defined as the change in internal energy of a system. The internal energy of a system is the sum of its kinetic and potential energies. The change in internal energy can be calculated using the equation:

\(\Delta U = q + w\)

where \(q\) is the heat added to or removed from the system, and \(w\) is the work done on or by the system. In the case of a bomb calorimeter, the system is the reaction taking place inside the calorimeter, and \(w\) is assumed to be zero since no work is done. Therefore, the equation simplifies to:

\(\Delta U = q\)

Given that the heat of combustion (\(q\)) of 2-propanol is -33.41 kJ/g, we can calculate \(\Delta U\) for the combustion of one mole of 2-propanol. The molar mass of 2-propanol (C3H8O) is:

\(Molar\ mass = 3 \times Atomic\ mass(C) + 8 \times Atomic\ mass(H) + 1 \times Atomic\ mass(O)\)

By looking up the atomic masses of carbon, hydrogen, and oxygen, we can calculate the molar mass.

\(Molar\ mass = 3 \times 12.01\ g/mol + 8 \times 1.01\ g/mol + 1 \times 16.00\ g/mol\)

Now we can calculate \(\Delta U\) for the combustion of one mole of 2-propanol using the equation:

\(\Delta U = q/Molar\ mass\)

Substituting the given value of \(q\) and the calculated value of the molar mass will give us the answer to part (a).

(b) \(\Delta H\) is defined as the change in enthalpy of a system. Enthalpy is a state function that depends on both the internal energy of the system and the pressure-volume work (\(PV\)) done by or on the system. The enthalpy change can be calculated using the equation:

\(\Delta H = \Delta U + P\Delta V\)

where \(\Delta U\) is the change in internal energy, \(P\) is the pressure, and \(\Delta V\) is the change in volume. In the case of a bomb calorimeter, the reaction takes place under constant volume conditions (\(\Delta V = 0\)), so the equation simplifies to:

\(\Delta H = \Delta U\)

Therefore, for the combustion of one mole of 2-propanol, the value of \(\Delta H\) will be the same as the value of \(\Delta U\), which we calculated in part (a).

To summarize:

(a) \(\Delta U = q/Molar\ mass\)
(b) \(\Delta H = \Delta U\)